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Program for North American Mobility in Higher Education Introducing Process Integration for Environmental Control in Engineering Curricula Module 4: Tier.

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Presentation on theme: "Program for North American Mobility in Higher Education Introducing Process Integration for Environmental Control in Engineering Curricula Module 4: Tier."— Presentation transcript:

1 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

2 2 LEGEND Go to the web site Go to next subject More information on the same subject Look for the answer to the question

3 Tier III: Open-Ended Problem

4 4 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.

5 5 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.

6 6 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.

7 7 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.

8 8 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.

9 9 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.

10 10 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.

11 11 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.

12 12 Necessary documents  LCA Base Case  Process Simulation

13 13 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.

14 14 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 Consumption23% reduction Liquid EffluentReduction of ion content of 2.3% Gas EffluentCl, K: 0.2% increase Na: 6.8% increase Energy produced5% reduction (need more energy to pump) Dust13.4% increase Solid wastesNeglictible difference

15 15 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.

16 16 Question 2 (Cont’d) Resource depletion:Emissions: ImpactWeight Raw water consumption 0.83 Energy consumption 0.08 Virgin Fiber consumption 0.01 Other resources 0.08 ImpactWeight Carcinogenic substances 0.70 Heavy metals0.07 Acidification0.01 Eutrophication0.01 Summer smog0.07 Winter smog0.07 Solid Wastes0.005 Global Warming 0.065

17 17 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

18 18 Solution – Q1 (Cont’d)  Water balances can be summarized by this picture.  The total fresh water consumption is = 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%).

19 19 Solution – Q1 (Cont’d)  The following graph show the contribution of each process unit to resource consumption.

20 20 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.

21 21 Solution – Q1 (Cont’d)  The following graph show the contribution of each process unit to emission-related environmental impacts.

22 22 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.

23 23 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.

24 24 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.

25 25 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.

26 26 Solution - Q2

27 27 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.

28 28 Solution - Q2 (Cont’d)

29 29 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.

30 30 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.

31 31 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

32 32 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. Weight of the Energy Aggregated Indicator

33 33 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

34 34 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

35 35 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.

36 36 Problem Statement – Q3-7  Consider the following Kraft pulp mill depicted below slaker green liquor clarifier white liquor clarifier causticizer Digester concentrator Recovery Boiler ESP Multiple Effect Evaporators grits dregs lime kiln cond. screening Brown Stock Washing chips white liquor black liquor weak pulp wash water dissolving tank mud washer mud filter dregs washer & filter weak white liquor dust recycle smelt salt cake SBL Flue Gas wash water wash water D E D E D fluegas lime mud To papermaking

37 37 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)

38 38 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.

39 39 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

40 40 Question 3  What are the rigorous targets in water discharge and reduction?

41 41 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.

42 42 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

43 43 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:

44 44 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

45 45 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).

46 46  With this knowledge, the ratio model can be used to relate the pollutant loads in streams 2 and 3 : P 3 new = (P 3 old / P 2 old ) * P 2 new  The pollutant load in stream 4 can then be determined by a simple component material balance: P 4 new = P 1 new + P 1 new + P 3 new Mathematical Modeling

47 47 Nominal Balance Model  By using these modeling techniques, path equations can be developed for tracking water and targeted NPE’s throughout the process, resulting in a mathematical model for the nominal case study. The nominal case study can then be revised to reflect the impact of mass integration on the process.

48 48 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

49 49 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

50 Digester Washer Screening MEEConcent. White Liq Clar Causticizer Slaker Lime Kiln Washers/ Filters Recovery Furnace Dissol. Tank Green Liq Clarifier ESP smelt Na 2 SO W1 = 3000 C1 = K1 = N1 = W2 = C2 = K2 = N2 = W4= C4 = K4 = N4 = W6 = 1450 C6 = K6 = N6 = W3 = 5127 C3 = K3 = N3 = W31 = 6143 C31 = K31 = N31 = W30 = 6143 C30 = K30 = N30 = W20 = 6402 C20 = K20 = N20 = W17 = 0 C17 = K17 = N17 = W16 = 0 C16 = K16 = N16 = W18 = 0 C18 = K18 = N18 = W26 = 423 C26 = K26 = N26 = W10 = 8901 C10 = K10 = N10 = W12 = 1024 C1 2= K12 = N12 = W14 = 0 C14 = K14 = N14 = W15 = 1202 C15 = K15 = N15 = W23 = 3.84 C23 = K23 = N23 = W19 = 6402 C19 = K19 = N19 = W5 = C5 = K5 = N5 = W32 = 1016 C32 = K32 = N32 = W27 = 0 C27 = K27 = N27 = W24 = 5762 C24 = K24 = N24 = W29 = 32 C29 = K29 = N29 = W28 = 8 C28 = K28 = N28 = W7 = C7 = K7 = N7 = W8 = 1450 C8 = K8 = N8 = W9 = 2225 C9 = K9 = N9 = W22 = 51 C22 = K22 = N22 = W25 = 423 C25 = K25 = N25 = W13 = 1202 C13 = K13 = N13 = W11 = 1202 C11 = K11 = N11 = W21= 6351 C21= K21 = N21 = Stripper Wood chips Pulp Bleach Plant W35 = Bleached pulp to papermaking 33 W33 = W37 = C37 = K37 = N37 = Chemicals Material Balance

51 51 Back to Question 3  What are the rigorous targets in water discharge and reduction? The objective here is to minimize the amount of fresh water used in the process and the amount of wastewater discharged from the system.

52 52 Solution – Q3  Beginning with the nominal balance model (figure ), the first step is to identify all possible sources of water entering, leaving or being consumed in the process in order to obtain the overall water balance for the process, as depicted in next figure

53 53 Bleach Plant = Chips = 3000 Washer = 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 = Pulp leaving Bleach Plant = OVERALL Water Balance Water consumed By reaction = 168 Total Water In = – 168 = tpd Total Water Out = tpd OVERALL WATER BALANCE

54 54 Solution – Q3  Next, all streams that use fresh water and all streams that contain potentially recyclable water are identified. There are four fresh water streams (S2, S6, S24 and S34) giving a total fresh water use of 52,197tpd. There are also four potentially recyclable streams, S8, S10, S12 and S37 giving a total of 42,365 tpd.  The overall water balance diagram has been modified to reflect this information (see figure )

55 55 Washer = Screening = 1450 Washers/Filters = 5762 Screening = 1450 MEE = 8901 Stripper = 1024 BP water = Water Balance Water consumed By reaction = 168 Total Fresh Water in = tpd Total recyclable water out = tpd Bleach Plant = FRESH AND RECYCLABLE WATER BALANCE

56 56 Solution – Q3  If the recyclable water can be intercepted and cleaned up to the point where it is acceptable for use in place of fresh water and if self-recycle is allowed, then one can determine the target for fresh water usage: Minimum water consumption = – = 9832 tpd

57 57 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

58 58 Water Consumed 168 WtoBP = W1 = 3000 W2 = W6 = 1450 W24 = 5762 Water going out with pulp tpd Wastewater Target = 1669 tpd Fresh Water Target = 9832 Chips Target for minimum water consumption = 52,197 – 42,365 = 9,832 tons per day OVERALL WATER TARGETING FOR CASE STUDY

59 59 Solution – Q3  Therefore, the rigorous targets are: Minimum Fresh water target = 9832 tpd Wastewater target = 1669 tpd

60 60 Limitations on Self-Recycle  Previously, it was permitted to consider recycling the effluent back to the same unit. However, self-recycle may sometimes be forbidden due to numerous reasons such as: To prevent the build-up of impurities in a flow loop To avoid dynamic instabilities that may arise due to the high interconnectivity between the input and output To enhance process reliability by disengaging the dependence of the input from the output.

61 61 Limitations on Self-Recycle  If self-recycle is not allowed, then it is possible that the targets identified earlier may not be reached even if interception technologies are used to clean up the recyclable water streams. As a result, new targets will need to be determined, which leads to the next question:

62 62 Question 4a  In the case of no self-recycle with one interceptor, which streams can be intercepted?

63 63 Solution Q4a  There are four recyclable streams for consideration: W8 – MEE W10 – Concentrator W12 – Screen W37 – Bleach plant effluent  In the development of the nominal balance model, it was assumed that there were no ions in the water leaving the MEE and Concentrator (i.e. it has the same quality as demineralized water); therefore, the only interception candidates are the screen and bleach plant effluents.

64 64 Question 4b  Choosing the bleach plant effluent for interception and assuming that the quality of the screen effluent is acceptable for direct recycle to the pulping process, what are the new water targets (remember, no self- recycle)?

65 65 Solution Q4b  The flow of the intercepted bleach plant effluent, along with the screen effluent is more than enough to replace all of the fresh water used in the pulping process. Therefore, the fresh water target for the pulping process is zero.  For the bleach plant, only water meeting dimineralized quality can be used. Thus, the effluents from the multiple effect evaporator and the concentrators can be used, replacing a total of 9925 tpd of BPE.

66 66 Solution – Q4b Bleach PULPING Consumption by Chemical Reaction and other losses = 9832 tpd Fresh water = tpd W8 = 1450 W10 = 8901 W12 = 1024 W2 = W6 = 1450 W24 = 5762 Wastewater To bio =30990 – = tpd = interception

67 67 Solution Q4b  The new targets are now: Pulping process fresh water: 0 tpd Bleach Plant Effluent fresh water target: – ( ) = tpd Wastewater target: – = tpd

68 68 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

69 69 Why Process Integration?  Process Integration is a holistic approach to the design and operation of complex systems. It is a sound framework that utilizes well-developed and proven mass and energy integration techniques for optimizing the design and operation of a process.

70 70 Process Integration  It is important to coordinate both process integration and process simulation. The application of process integration provides performance targets, solution strategies, and proposed changes to the process. Process simulation reassesses the process performance as a result of theses changes.

71 71 Coordination of Process Integration and Simulation Process Integration Process Simulation 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 relations New Processes Process Modifications Structural changes

72 72 Mass Integration Strategies  Now that the rigorous targets have been developed for the minimum feasible water usage and discharge, various cost-effective mass integration strategies should be used to attain the targets. These strategies include Segregation, Low cost/no cost modifications, Direct recycle, Interception high cost process modifications.  The above strategies can be represented as a pyramid (see next slide), where it is desired to begin at the bottom of the pyramid, which represents the lowest cost and perhaps more easily implemented strategies, and work up until the target is achieved.

73 73 Mixing & Recycling Interception Segregation Target Chemical Process Low Cost Process Modifications (LCPM) HCPM Mass Integration Strategies

74 74 Segregation  Segregation refers to avoiding the mixing of streams. In some industrial applications, dilute streams have been mixed with concentrated streams and even different phases have been mixed together unnecessarily. Segregation of streams at the sources can provide several opportunities for cost reduction such as: Generate environmentally benign streams Enhance the opportunities for direct recycle since dilute streams are easier to recycle. The separate concentrated streams are now more thermodynamically favorable for interception

75 75 Low-cost process modifications  In some cases, a change in process conditions (such as temperature, pressure, compositions, etc) may be all this is needed to decrease or eliminate the waste produced in a unit.  Provided that the cost is low, a unit can be replaced with a more environmentally benign one.

76 76 Recycle  Discharged waste can be reduced by recycling pollutant-laden streams back to the process to be utilized in process or non-process requirements. In some instances, several streams need to be mixed with each other to achieve the desired level of flowrate and composition.

77 77 Interception  Interception refers to the utilization of separation techniques to selectively remove targeted species from targeted streams. In most industrial applications, inteception is needed to enhance the opportunities of recycling and to generate environmentally benign streams.

78 78 High-Cost Process Modifications  After all other strategies have been exhausted, one may need to employ high cost process modifications. This may include completely new chemistry (such as new solvent or new reaction path), new technology (new plant), etc.

79 79 Question 5  What is the optimal water allocation using direct recycle?

80 80 Solution – Q5  To answer this question, a mass allocation representation of the process from the species viewpoint needs to be developed.  For each species, there are sources, those streams that contain the desired species, and sinks, those streams units which can accept the species.  Each sources can be segregated, intercepted to adjust species content, mixed, etc and allocated to the different units or sinks, as depicted in the following figure.

81 81 Source i = 1 Fresh Source Segregated Sources Sinks Species Interception Network (SPIN) j = 1 j = 2 j = N sinks Source i = N sources 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 82 Process Sinks  There are a number of process units, or N sinks,that employ fresh water and are designated by the index j (j ranges from 1 to N sinks ).  Each j th sink has two sets of contraints on flowrates and composition: Flowrate to each sink W j min  W j  W j max j = 1, 2,….,N sinks W j is the water flowrate entering the j th sink Ion content to each sink Y ion,j min  Y ion,j,  Y ion,j max j = 1, 2,….,N sinks Y ion,j is the compostion of a certain NPE entering the j th sink

83 83 Process Sinks  Each source, represented by I, is split into N sink fractions that can be assigned to various sinks. The flowrate of each split is denoted by l i,j (see figure)  Each split fraction then has the opportunity to be mixed (or not) and assigned to sinks (see figure)

84 84 Splitting of the i th source: where i = 1,2, …, N sources Source L i Y ion,i l i,j SPLITTING OF SOURCES TO SINKS

85 85 Mixing for the j th sink: where j = 1,2, …, N sinks j l i,j y ion,j W j Y ion,j MIXING OF SOURCES BEFORE SINKS

86 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 87 Sinks Screening Brown Stock Washer Washers/Filters Bleach Plant Sources S 8 Wastewater from Screening Fresh water S 10 Condensate from MEE S 12 Condensate fromConcenrator S 37 Bleach Plant Effluent Direct Recycle/Reuse Representation Biotreatment

88 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

89 89 Direct Recycle Optimization Formulation for Source/Sink Analysis w/Path Connection j = 1, 2, …, N sinks NPE content in feed to each sink: j = 1, 2, …, N sinks and k = 1, 2, …, N k Splitting for the i th source: i = 1, 2, …, N sources Mixing for the j th sink: j = 1, 2, …, N sinks Component material balances for the pollutants: j = 1, 2, …, N sinks and k = 1, 2, …, N k Flowrate to each sink: Non-negativity of each fraction of split sources: i = 1, 2, …, N sources and j = 1, 2, …, N sinks

90 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 C i, N i, and K i, are the ionic loads of Cl, Na and K, respectively, in the i th source:

91 91 Optimization Solution for Direct Recycle/Reuse  LINGO programming was used to develop and solve the mathematical formulation. The optimal water allocation is depicted in the following slide. The fresh water to screening has been replaced with 751 tpd of concentrator effluent and 699 tpd of bleach plant effluent The fresh water to the washers/filters has been replaced with 273 tpd of concentrator effluent and 5489 tpd of MEE effluent. A portion of the fresh water to the Brown Stock Washers has been replaced with 3412 tpd of MEE effluent and 1450 tpd of screening effluent.

92 92 White Liq. Clarif. Causticizer Screening MEE Concent. Lime Kiln Washers/ Filters Slaker Dissolv Tank Green Liq. Clarif. Recovery Furnace ESP BSW Digester Stripper Wood Chips Stripper Bleach Optimum Solution for Direct Recycle/Ruse

93 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 94 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?

95 95 Solution – Q6  In this problem, the objective function has changed from one of minimizing the fresh water consumption to one of minimizing the load of the Cl to be removed from the bleach plant effluent subject to: Desired water target Path equations for tracking water and Cl Recycle Model Interception equations Unit constraints

96 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 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).

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

98 98 Exploring other interception opportunities  So far, only terminal streams (those streams going directly to waste treatment) have been considered. However, it is possible that other inter-process streams may be intercepted, perhaps providing greater economical and environmental benefits.  A literature search reveals that salt removal technologies exist for other kraft units, among those: White Liquor Interception Green Liquor Interception  Of course, this leads to the next question:

99 99 Question 7  How much chloride needs to be removed from Case 1: Green Liquor Case 2: White Liquor in order to meet the fresh water target?

100 100 Interception Alternatives  This is quick and easy to determine. The objective function will remain the same (minimize chloride removal) as in Q6 but rather than minimizing the Cl removal from the bleach plant effluent, it will be minimized from the white liquor or green liquor streams. Thus, the optimization program only needs to be slightly altered to reflect the stream in question.  Interestingly enough, though it should come as no surprise, the load removal for Green Liquor and White Liquor interception is the same as the case for Bleach Plant interception (approx. 9 tpd of Cl). However, the three solutions are not identical. Each one has a different configuration of optimal water allocation (see figures)

101 101 White Liq. Clarif. Causticizer Screening MEE Concent. Lime Kiln Washers/ Filters Slaker Dissolv Tank Green Liq. Clarif. Recovery Furnace ESP BSW Digester Stripper Wood Chips Stripper Bleach Optimum Solution for Bleach Plant Interception

102 102 White Liq. Clarif. Causticizer Screening MEE Concent. Lime Kiln Washers/ Filters Slaker Dissolv Tank Green Liq. Clarif. Recovery Furnace ESP BSW Digester Stripper Wood Chips Stripper Bleach Optimum Solution for Green Liquor Interception

103 103 White Liq. Clarif. Causticizer Screening MEE Concent. Lime Kiln Washers/ Filters Slaker Dissolv Tank Green Liq. Clarif. Recovery Furnace ESP BSW Digester Stripper Wood Chips Stripper Bleach Optimum Solution for White Liquor Interception

104 104 Life Cycle Analysis  But which of the three technologies is the better solution?

105 END

106 106 Path Diagram Equation  Typically, the Path Diagram Equation defines outlet flows and compositions from key units as functions of inlet flows, inlet compositions and process design and operating conditions  This mass integration tool tracks the targeted species as they propagate through the system and provide the right level of details that will be incorporated into the mass integration analysis Return to the flowsheet

107 107 1.Degrees of Freedom N V = N S x N C F= N V - N E = N C (N S - 1) F: degrees of freedom N V : number of variables N E : number of equations N C : number of targeted species N S : 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 streams N streams out Return to the flowsheet

108 108 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

109 109 Fixed Split Flow Model Fixed Split Model F 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

110 110 Flow Ratio Model Flow Ratio Model F G G new = G old * ( F new / F old )  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

111 111 Species Ratio Model Species Ratio Model F G II new = I new (II old / I old )  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 1 II = species 2 Return to the flowsheet

112 112 Initial Data - Digester  Assumption: all inlet streams are known.  Flowrate of wood chips, Chips = 6000 tpd  Moisture content of wood chips = 50%  Pulp Yield = 50%  Pulp = Dry Chips * Yield  Mass fraction ions with incoming wood chips: C 1 = 1 * Chips/6000 K 1 = 2.50 * Chips/6000 N 1 = * Chips/6000 Return to the flowsheet

113 113 Initial Data – Brown Stock Washer  Composition of ions in incoming wash water: Cl = 3.7 ppm K = 1.1 ppm Na = 3.6 ppm  Consistency of pulp leaving Brown Stock Washer, CY = 0.12  Dilution Factor, DF = 2.0  Ratio of ions in slurry leaving the BSW to the chloride in the pulp stream leaving the digester Cl = K = Na = Return to the flowsheet

114 114 Digester Brown Stock Washer S1S1 S2S2 S4S4 S5S5 S3S3 W 1 (from moisture content C 1 (from comp of Cl in chips) K 1 (from comp of K in chips N 1 (from comp of N in chips W 2 (from consistency) C 2 (from comp of Cl in wash water) K 2 (from comp of K in wash water) N 2 (from comp of N in wash water) All species data will be calculated as an output stream from white liquor clarifier W 4 (from dilution factor C 4 (from ratio to C 5 ) = 0.05*C 5 K 4 (from ratio to K 5 ) = 0.02*K 5 N 4 (from ratio to N 5 ) = 0.009*N 5 W5W5 C5C5 K5K5 N5N5

115 115 Digester W 1 = Moisture*Chips=0.5*6000=3000 W 4 = [(1-CY)/CY]*Pulp=[(1-.82)/(0.82)]*3000 DF = (W 2 - W 4 )*Pulp; DF is given as 2 W 2 can be determined from after W 4 has been calculated. Ion Content in streams 2 and 4: C 2 = (3.7*10 -6 ) *W 2 ;C 4 = 0.05*C 5 K 2 = (1.1*10 -6 ) *W 2 ;K 4 = 0.02*K 5 N 2 = (3.6*10 -6 ) *W 2 ;N 4 = 0.009*N 5 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 S 5 ); these can be obtained Via the four material balances for the 4 species W 5 = W 1 + W 2 + W 3 – W 4 C 5 = C 1 + C 2 + C 3 – C 4 K 5 = K 1 + C 2 + K 3 – K 4 N 5 = N 1 + C 2 + N 3 – N 4

116 116 Multiple Effect Evaporator  80% of the water in the weak black liquor is evaporated (water recovery ratio is 0.8).  It is assumed that no ions are in the condensate of the multiple effect evaporators  The material balances can be used to calculated the concentrated stream leaving the multiple effect evaporators W 10 = Water recovery in evaporator * W 5 W 9 = W 5 - W 10 C 9 = C 5 - C 10 K 9 = K 5 - K 10 N 9 = N 5 - N 10

117 117 Multiple Effect Evaporators Multiple Effect Evaporators W 10 = Water Recovery * W 5 C 10 = 0 K 10 = 0 N 10 = 0 S5S5 W5W5 C5C5 K5K5 N5N5 Black Liquor S 10 Evaporator Condensate S9S9 Evaporator Concentrate W9W9 C9C9 K9K9 N9N9

118 118 Multiple Effect Evaporator  46% of the water in the black liquor entering the concentrators is evaporated (water recovery ratio is 0.46).  Again, it is assumed that no ions are in the condensate of the multiple effect evaporators  The material balances can be used to calculated the concentrated stream leaving the multiple effect evaporators W 12 = Water recovery in concentrator * W 9 W 11 = W 9 - W 12 C 11 = C 9 - C 12 K 11 = K 9 - K 12 N 11 = N 9 - N 12

119 119 Concentrator W 11 C 11 K 11 N 11 S9S9 Evaporator Concentrate W9W9 C9C9 K9K9 N9N9 S 10 Concentrator Condensate W 12 = Water Recovery * W 9 C 12 = 0 K 12 = 0 N 12 = 0 S 11 Strong Black Liquor

120 120 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 W 15 = W 11.  The ions in the solids return, ESP dust and off-gass are related to the ions in the strong black liquor stream: C 13 = 0.278*C 11 ;C 14 = 0.048*C 11 ;C 15 = 0.02*C 11 K 13 = 0.498*K 2 ;K 14 = 0.028*K 11 ;K 15 = 0.008*K 11 N 13 = 0.154*N 2 ;N 14 = 0.002*N 11 ;N 15 = *N 11

121 121 Recovery Furnace and ESP  The component material balance around the ESP is: W 13 - W 14 - W 15 - W 16 = 0.0 C 13 - C 14 - C 15 - C 16 = 0.0 K 13 - K 14 - K 15 - K 16 = 0.0 N 13 - N 14 - N 15 - N 16 = 0.0  It is assumed that the saltcake has a makeup flow of * Pulp. Knowing this and the molecular formula for saltcake, N 18 = 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. W 17 = 0.0 W 18 = 0.0 C 18 = 0.01*N 18 K 18 = *N 18

122 122 Recovery Furnace and ESP  The ion content in the smelt is determined via component material balance around the Recovery Furnace and ESP C 11 + C 18 - C 15 - C 14 - C 17 = 0.0 K 11 + K 18 - K 15 - K 14 - K 17 = 0.0 N 11 + N 18 - N 15 - N 14 - N 17 = 0.0

123 123 Smelt  The smelt flowrate consists of the saltcake + solids in strong black liquor (SBL) – solids lost with the purge streams (S 14 and S 15 ) – 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

124 124 Recovery Furnace and Electrostatic Precipitator (ESP Recovery Furnace ESP W 15 = W 11 C 15 (from ratio to C 11 ) K 15 (from ratio to K 11 ) N 15 (from ratio to N 11 ) W 18 = 0 C 18 (from ratio to N 18 ) K 18 (from ratio to N 18 ) N 18 = * Salt Cake W 14 = 0 C 14 (from ratio to C 11 ) K 14 (from ratio to K 11 ) N 14 (from ratio to N 11 ) W 17 C 17 K 17 N 17 S 17 S 14 S 18 S 15 S 11 Strong Black Liquor W 11 C 11 K 11 N 11 Smelt Dust Purge Off-gas Salt Cake = *Pulp

125 125 Dissolving Tank  The dissolving water-to-smelt ratio used in the dissolving tank is typically 85/15 W 19 = (85/15)*Smelt  The ionic content of S 19 is determined by assuming ratios of Cl and K to those in the smelt and Na to the white liquor C 19 =0.136*C 17 K 19 =0.136*K 17 N 19 =0.196*N 3  Component material balances around the dissolving tank are used to evaluate the ionic content of the feed to the green liquor clarifier W 20 - W 17 - W 19 = 0.0 C 20 - C 17 - C 19 = 0.0 K 20 - K 17 - K 19 = 0.0 N 20 - N 17 - N 19 = 0.0

126 126 Dissolving Tank Dissolving Tank W 17 C 17 K 17 N 17 S 17 Smelt W 19 = 5.67 * Smelt C 19 (from ratio to C 17 ) K 19 (from ratio to K 17 ) N 19 (from ratio to N 17 ) S 19 Dissolving Water = (85/15)*Smelt S 20 Feed to green liquor clarifier W 20 C 20 K 20 N 20

127 127 Green Liquor Clarifier  Typical overflow ratios were used to obtain the flows and ion concentrations in the overflow and underflow stream. W 21 =0.992*W 20 C 21 =0.863*C 20 K 21 =0.880*K 20 N 21 =0.968*N 20  Component material balance can then be written around the green liquor clarifier: W 22 - W 21 - W 20 = 0.0 C 22 - C 21 - C 20 = 0.0 K 22 – K 21 - K 20 = 0.0 N 20 - N 21 - N 20 = 0.0

128 128 Green Liquor Clarifier Green Liquor Clarifier S 20 Feed to green liquor clarifier W 20 C 20 K 20 N 20 C 21 (from ratio to C 20 ) K 21 (from ratio to K 20 ) N 21 (from ratio to N 20 ) S 21 Dust Overflow W 21 (from ratio to W 20 ) S 22 Underflow W 22 C 22 K 22 N 22

129 129 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 dregs W 23 = 0.075*W 22 C 23 = 0.010*C 22 K 23 = 0.001*K 22 N 23 = 0.250*N 22  The overflow from the white liquor clarifier is determined by assuming a ration to Green Liquor overflow W 32 = 0.160*W 21 C 32 = 0.237*C 21 K 32 = 0.016*K 21 N 32 = 0.156*N 21

130 130 Washer/Filter System  The wash water is assumed to be 90% of the smelt dissolution water and the ionic content for C, K and N is based on the typical values of 3.7, 1.1, and 3.6 ppm, respectively W 24 = 0.9*W 19 C 24 = (3.7*10 -6 ) *W 24 K 24 = (1.1*10 -6 ) *W 24 N 24 = (3.6*10 -6 ) *W 24  Component material balances can then be written for the washer/filter system W 22 + W 24 + W 32 - W 19 - W 23 - W 25 = 0.0 C 22 + C 24 + C 32 - C 19 - C 23 - C 25 = 0.0 K 22 + K 24 + K 32 - K 19 - K 23 - K 25 = 0.0 N 22 + N 24 + N 32 - N 19 - N 23 - N 25 = 0.0

131 131 Washer/Filter S 22 Underflow W 22 C 22 K 22 N 22 S 23 Dregs S 19 To dissolving tank S 25 Feed to lime kiln W 25 C25C25 K 25 N 25 C 32 (from ratio to C 21 ) K 32 (from ratio to K 21 ) N 32 (from ratio to N 21 ) S 32 Overflow from WLC W 32 (from ratio to W 21 ) S 24 Wash Water Ion content of Cl, K and Na is assumed to be 3.7, 1.1 and 3.6 ppm

132 132 Lime Kiln  The lime leaving the lime kiln is assumed to have no water. It is assumed that 95% of the sodium entering the lime kiln leaves in the off-gas. The ratio of C and K to water in the off-gas is assumed to be W 27 = 0.0 W 25 = W 26 C 26 = *W 26 K 26 = *W 26 N 24 = 0.05 *N 25  A material balance can then be written around the kiln C 25 – C 26 – C 27 = 0 K 25 – K 26 – K 27 = 0 N 25 – N 26 – N 27 = 0

133 133 Lime Kiln S 27 To Slaker W 27 C27C27 K 27 N 27 S 25 Lime mud from washer/filter W 25 C25C25 K 25 N 25 S 26 Kiln Off-gas W 26 C26C26 K 26 N 26

134 134 Slaker  The slaking reaction is given by CaO + H 2 O = 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 W 29 = * W 21 C 296 = 0.0 K 296 = 0.0 N 294 = 0.0

135 135 Slaker  The ion content in the grits is related to the green liquor overflow W 28 = * W 21 C 28 = *C 21 K 28 = *K 21 N 28 = *N 215  The component material balance can then be written around the slaker: W 21 + W 27 - W 28 - W 29 - W 30 - Waterslk = 0.0 C 21 + C 27 - C 28 - C 29 - C 30 = 0.0 K 21 + K 27 - K 28 - K 29 - K 30 = 0.0 N 21 + N 27 - N 28 - N 29 - N 30 = 0.0

136 136 Slaker S 30 To Causticizer S 27 From Lime Kiln S 21 Green Liquor Overflow S 29 Slaker Vapor S 28 Grits C 29 = 0 K 29 = 0 N 29 = 0 W 29 = *W 21 C 21 (from ratio to C 21 ) K 21 (from ratio to K 21 ) N 21 (from ratio to N 21 ) W 21 (from ratio to W 21 ) Water consumed by Rxn 0.032*Lime = 0.35*Pulp

137 137 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) W 31 = W 30 C 31 = C 30 K 31 = K 30 N 31 = N 30  The material balance around the WLC is then W 21 - W 32 - W 3 = 0.0 C 21 - C 32 - C 2 = 0.0 K 21 – K 32 - K 3 = 0.0 N 21 - N 32 - N 3 = 0.0

138 138 Causticizer and White Liquor Clarifier White Liquor Clarifier Causticizer S 30 From Slaker S 31 White Liquor to digester S 32 To washer/filter

139 139 Bleaching

140 140 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.

141 141 Degree of Freedom Analysis Unit U INLET STREAMS (fresh inputs or outlets from other units) OUTLET STREAMS NSNS Given the following generic unit and F = Number of Degrees of Freedom N V = number of unknown variables (N S *N C ) N E = number of equations in each variable N S = number of outlet streams N C = number of targeted species

142 142 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 = N V – N E = N C (N S – 1)


144 144 Screen Brown Stock Washer Washer/Filter Bleach Plant Biotreatment Fresh 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 f101 u = 1 u = 2 u = 3 u = 4 u = 5 f102 fresh4 f121 f103 f105 f125 f85 f37bio f371f104 fresh1 f82 f83 fresh3 fresh2 f122 f123 f124 f81 f83 f84 f373 f372 Optimization of Source/Sink Analysis With Path Connections

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