Letter of Transmittal s2

PAPERNOL Team University of Tennessee at Chattanooga 615 McCallie Avenue Chattanooga, TN 37403-2598 10 April 2008

Dr. Jim Henry Professor, P.E. University of Tennessee at Chattanooga College of Engineering & Computer Science Dept. 2402, 435 EMCS 615 McCallie Avenue Chattanooga, TN 37403-2598

Dr. Henry:

The following report details the Papernol design process. Papernol is an attempt to convert waste cellulose from a paper source to fuel grade ethanol. The goal is to present a viable and comprehensive explanation of how to achieve this deliverable. The specifically designed process includes five major steps plus utilities. These steps are preparation, hydrolysis, evaporation, fermentation, and separation.

The report contains the following information:  Introduction to system  Full description and design specifications of each step  Analysis of energy requirements  Conclusive summary of the designed Papernol process

Thank you for your time.

Sincerely,

Laura Amini, Dianah Dugan, Jason Frizzell, Taylor Murphy, Josh Oliver, Alicia Redd, and Bryan Varnell PAPERNOL Technical, Environmental, and Economic Feasibility Study of Production of Ethanol from Recycled Paper

UTC Senior Engineering Design Project 24 April 2008

University of Tennessee at Chattanooga College of Engineering and Computer Science ENCH 430 615 McCallie Avenue Chattanooga, TN 37403

PAPERNOL Team Members Laura Amini Dianah Dugan Jason Frizzell Taylor Murphy Josh Oliver Alicia Redd Bryan Varnell Executive Summary

The Papernol Design Group has been tasked to develop a process, dubbed the “Papernol” process, which converts cellulosic material found in waste paper into ethanol. The project goal is to design a plant based on the current demands of the Orange Grove Recycling facility. This facility obtains 10,000 pounds of waste paper per day. The Papernol plant will produce ethanol as a competitive liquid fuel alternative to gasoline using Orange Grove’s waste paper feedstock. The team has developed a preliminary Project Statement outlining the major goals of the project.

The Papernol plant has been designed with six key sections: preparation, acid hydrolysis, evaporation, fermentation, separation, and system utilities. Although each of these sections is pertinent to the conversion of waste paper to ethanol, the system is economically evaluated with and without the presence of the evaporation system.

Preparation of the paper is carried out with a shredder in order to break the paper products into smaller fragments. The shredded paper products then are added with water to form slurry (7% paper by mass). The slurry mixture is essential in more effectively converting into glucose from the cellulose by the digestion system.

Digestion will be performed using dilute acid hydrolysis, a process developed by TVA in the 1950s, for the saccharification of the cellulose. The hydrolysis will produce a solution of simple sugars, predominantly glucose, from the cellulose in the paper. This simple sugar solution will be the feedstock for the fermentation vessel [2].

Once the acid hydrolysis section has successfully produced a glucose solution from the paper, an evaporator will then reduce the amount of water in the solution, thus, increasing the sugar concentration for optimal results in the fermentation.

The fermentation vessel will be utilized to convert the simple sugar solution into ethanol and water via fungal and/or bacterial fermentation. The fermentation process will be initiated by inoculating the sugar solution with a commercial yeast amendment. One bi- product of ethanol production using yeast is the formation of Carbon Dioxide (CO2). This will be bottled upon release [2].

The distillation column is the separation process subsystem. Distillation will separate the ethanol from the mixed ethanol-water solution fed to the distillation column from the fermentation subsystem. This distillation feedstock will consist of the ethanol-water mixture produced by the yeast fermentation [2].

The Utilities section will consist of three main components: the cooling water tower, reboiler, and electricity. The cooling water tower is responsible for supplying the required amounts of water to each of the five sections. The reboiler will be run on natural gas and also supply low pressure steam to the sections needing steam. And electricity is mainly supplied for the pumps within the systems. Utilities section will also analyze the cost of the system per day and potential sources of minimizing these costs.

ii Throughout the process, material balances, energy balances, and sizing were all considered and calculated. Given this data, the 10,000 lbs of paper is converted through the Papernol system into 500 gallons of ethanol.

The economic analysis is over a ten year period with an estimated budget of 3 million dollars to build the system. The price of producing one gallon of ethanol is examined with the effects of a recycle stream and an evaporator. The final price for the Papernol system to produce one gallon of ethanol is $1.40.

iii Table of Contents Executive Summary ...... ii Table of Contents...... iv Problem Statement ...... 1 Introduction to System ...... 2 Section 100 – Preparation...... 5 Introduction ...... 5 Results ...... 6 Discussion...... 9 Section 200 – Hydrolysis ...... 11 Introduction...... 11 Results...... 12 Discussion...... 15 Section 300 – Evaporation...... 16 Introduction ...... 16 Results...... 19 Discussion...... 22 Section 400 – Fermentation ...... 23 Introduction ...... 23 Results ...... 27 Discussion...... 29 Section 500 – Separation...... 30 Introduction ...... 30 Results...... 31 Discussion...... 35 Section 600 – Utilities...... 36 Introduction ...... 36 Results...... 38 Discussion...... 42 Conclusions...... 43 Recommendations...... 47 References ...... 48 Appendices ...... 49

iv Problem Statement

The UTC - PAPERNOL design team has designed a plant to convert waste paper at 10,000 lbs per day to ethanol with 99% purity.

The design team has completed  essential mass and energy balances  block and process flow diagrams  technical, economic and environmental feasibilities of the process

1 Introduction to System

The purpose of the following process is to convert waste cellulose from a paper source into fuel grade ethanol. This process has been named Papernol by the design team since paper and ethanol are the keywords associated with this project and the transition from paper to ethanol is the primary goal. The ability to convert waste paper, or paper that is not currently recycled on a large basis, which contains cellulose into a useable product such as ethanol, is a valuable asset.

Currently, ethanol is being pursued as an alternative to gasoline for use as commercial fuel. In the United States alone, 388.6 million gallons of gasoline are consumed per day for use as fuel in transportation [1]. However, with fuel prices on the rise and a limit on the decreasing supply of fossil fuels, alternative sources of fuel must be supplied. Ethanol produced from biological sources is the primary solution being used to offset the growing gasoline demand with approximately 4 billion gallons produced in 2005 [2]. This bioethanol is created by converting cellulose in feedstocks such as corn and switchgrass into ethanol. Unfortunately, using crops as the source of ethanol requires farmland dedicated solely to the production of useable sugar source. This burden on farmland is just as limited as the supply of fossil fuels and as such could never replace the gasoline needs in the United States.

However, there is a source of cellulose circulating in the United States today that is not being used. Paper is used by the tons everyday. According to the Paper Industry Association Council, annual paper and paperboard consumption in the United States is 100.2 million tons. Approximately 53.4% of this material (53.5 million tons) is recovered each year. Additionally, approximately 33 million tons of corrugated containerboard were consumed in 2006. Approximately 76.4% of this or 25.2 million tons were recovered. Thus, about 78.7 million tons of recycled paper, paperboard, and container board are available as a potential cellulose feedstock [2]. This leaves tons of paper that is not recycled. Usually, it would end up in a landfill where it simply degrades back into cellulose and washes into the soil. The goal of Papernol is to take this unrecycled and otherwise undesired paper stock and use it to create ethanol.

This has several other advantages over the crop feedstocks mentioned earlier. It requires no farmland to grow the cellulose. Paper companies and paper consumers effectively cover this burden since they originally acquired and used the cellulose source. Another complaint against corn-based ethanol is that it seizes a potential source of food and nourishment from low income areas of the world and raises overall food prices in general. Papernol does not encounter this problem since no farmland is required. Furthermore, there is no labor and energy input required to grow the feedstock. Also, the corn crops are grown primarily in the farm belt of the United States and must be transported to the places with the highest demand for fuel after being converted to ethanol. This requires further energy and fuel needs to transport ethanol in tanker trucks long distances. This differs from paper since the areas with the largest fuel demand generally have large waste paper sources. Again, the paper companies and paper consumers have offset this transportation cost since the paper was transported to the

2 consumer for the initial purchase. This process turns a waste stream into a revenue stream.

This is achieved by a series of chemical reactions and separations collectively called the Papernol process. Each step has a specific goal and desired outcome. These steps will be discussed in detail in the corresponding section.

Figure 000-1. Block Diagram of the Entire Papernol Process

Figure 000-1 is the block diagram for the Papernol system. The first step is preparation. It is dubbed Section 100. This step consists of handling the paper, shredding, and mixing it with water to form slurry. The slurry is sent to Section 200. The hydrolysis section is responsible for heating the slurry, hydrolyzing the cellulose to glucose, and filtering out unwanted products before sending it to fermentation. Section 300 aims to remove some of the excess water through evaporation. This solution can then be sent to Section 400, fermentation. As one of the most important steps, fermentation turns glucose into the final product in the form of an ethanol mixture. For handling this mixture, it is sent to Section 500. This section is called separation. The separation is achieved through a distillation column and molecular sieves. The final product is then 99% pure ethanol. Section 600 is the utilities section and deals with providing each previous section with the necessary energy.

3 This brief overview just lists the basics of the Papernol process. In general, fuel grade ethanol is in high demand, and the Papernol process seeks to provide an alternate source of ethanol. This is achieved using waste paper as feedstock through a specially designed process.

4 Section 100 Introduction

The initial step in the Papernol process is preparation. Preparation consists of converting waste paper into slurry by adding water. There are two main components that make up this phase, shredding and mixing. The shredding is necessary to break the paper down into smaller pieces so that water may be added effortlessly. Mixing has to be the next stage in preparation so the transition to the other steps in the process can be carried out.

5 Section 100 Results

Material Balance

From preliminary research, the basis for the section was 10,000 pounds of waste paper per day. One assumption made in the preparation stage was that a truck would drop off the paper. This entails that the paper must be stored temporarily before being introduced into the actual Papernol process. Taking this into account, a storage hopper was considered as the storage device.

Also found through research, was the ratio of water to paper for the appropriate slurry to be made. This ratio is 15 and was found to be most suitable because it allows the slurry to move through each phase with the fewest difficulties (i.e. pumps and filters cannot handle any larger solid mass concentration than that). Knowing this ratio, the mass of water needed to be supplied during mixing was calculated and the material balances for both shredding and mixing were complete. Figure 100-1 illustrates these numbers and the appropriate equipment for each.

Hopper/ Mixer Shredd 10,000 lb paper 10,000 lb paper 160,000 lb slurry er

150,000 lb water

Figure 100-1: Mass Balance for Paper Shredder and Mixer

From Figure 100-1, it can be seen that the material balance for both the hopper and the shredder has the same amount of paper in and out. Since the hopper and shredder do not add any other components, the same amount that goes in must also come out of the device.

The mixer has to take the paper mass that comes from the shredder and add an amount of water according to the ratio discussed previously. This, as Figure 100-1 shows, results in the total mass leaving the mixer being 160,000 pounds of slurry. The mass balances represent the amount of slurry produced throughout one day, which corresponds to the system basis. Another assumption to be made for the preparation stage is that the average composition of waste paper that will be used. It was determined to be composed of 80

6 percent cellulose and 20 percent other waste. Using this number, the 160,000 pounds of slurry made during preparation contains 8,000 pounds of cellulose that will be fed to the hydrolysis vessel in Section 200.

Energy Balance and Equipment Sizing

The energy balance for the preparation phase has no temperature or pressure changes from inlet to outlet in any piece of equipment; therefore all energy in this phase will consist of electrical energy. The electrical energy will be calculated based on the amount of power required to operate the equipment. This leads to choosing the equipment and determining the size and cost for each component.

To determine the size of equipment, the mass coming in and going out for each must be considered. The following equation was used to determine the appropriate volume:

Q = mdot / ρ (100-1)

3 Where Q is volumetric flow rate in cubic feet per day (ft /day), mdot is the mass flow rate in pounds per day (lbs/day), and ρ is the density in lbs/ft3. The volumes for equipment in a continuous process also depends on residence time tr (hr) and this volumetric flow rate Q. This is shown in Equation 100-2.

V = tr * Q / 24 (100-2)

This equation relates the volumetric flow rate through the entire system to the time spent flowing through a single piece of equipment. This produces an accurate measure of the volume required for any equipment used in a continuous process. Through research, the density of paper was found to be approximated by .025 lb/in3 [3]. With this density substituted into Equations 100-1 and 100-2, the values for equipment volume were found. These values are shown in Table 100-1.

Equipment Volume (ft3) Storage Hopper 235 Shredder 77 Mixer 880 Table 100-1: Equipment Sizing

Equipment Cost

After the appropriate size was determined, the power and cost were considered. The best estimation of cost is to find a manufacture with the size requirements for the equipment and use the purchasing price. This was found by looking at different manufactures and comparing the average price of equipment. Using the material balance to find out the amount of paper needed to be processed within a day, the power was also determined the same way as cost. All values are displayed in Table 100-2.

7 Equipment Cost ($1000) Power (kW) Storage Hopper 2 0 Shredder 32 160 Mixer 42 56 Conveyors 22 1 Table 100-2: Cost and Power of Equipment

A piece of equipment not mentioned until this point, is the conveyor. For the preparation phase, there will be two different points that will need a conveyor. One will be used to unload the truck and convey the paper to the storage hopper. The other will be to convey the paper from the shredder to the mixer.

The price listed for the storage hopper includes multiple hoppers. Due to the mass of paper needed in the storage phase, there was not one hopper that could contain this quantity. The hoppers found to use for storage were approximately a third of the total mass, so the total number needed for preparation is three each having the price of approximately 650 dollars. Both the shredder and mixer have the capacity and power to perform their functions for the entire amount of mass needed. The mass of paper and water are large, therefore, the power and the price of the equipment are elevated. In order to rationalize the decision to spend this amount of money, another option was considered. This option was to have more than one shredder and mixer with less power requirement and more affordable purchasing prices. The downfall to this option would be the utility costs and time. To make 160,000 pounds of slurry, multiple (depending on size up to 4 of each) shredders and mixers would be used and in return the utility costs would rise. The time basis for preparation is a day, with less capacity capabilities; the slurry may not be ready within the time frame given for our basis. Because of the former problems of multiple shredders and mixers, the decision to spend more on purchasing cost, less on utility, and ensure the slurry is made in a day was justified and only one mixer and shredder would be used.

8 Section 100 Discussion

When researching and combining prior knowledge of technical aspects of the preparation phase, different options/obstacles arose. The previous decision about the shredder and mixer was only one.

The first problem encountered within this phase was how paper would be fed to the system. Originally the paper would be transferred from the truck to the shredder by a boom loader. After research, the boom loader was not feasible. The paper used for the process is loose paper and using a boom loader would allow the paper to be scattered when transporting. For this reason, a conveyor was thought to be more conducive. The conveyors chosen were trough and covered conveyors. Also to prevent paper being dispersed during conveying, enclosing the conveyor at this point in the phase is key. The trough conveyor will be used to transport paper from the shredder to the mixer. Using a trough shape will allow the paper to be semi concealed and the entire mass to be moved from one piece of equipment to the next.

Another problem was encountered when finding equipment pricing. When designing the original process flow diagram (PFD), mixing water and paper came before shredding. After attempting and failing to find a shredder that shreds mixtures, the decision to move shredding before introducing water into the system was made. Also if a shredder could shred mixtures, the purchasing and maintenance costs would be outside of the allotted budget for the process, which means it is not viable.

The entire preparation phase PFD is shown in Figure 100-2 below with stream and equipment information listed in Table 100-3 and 100-4.

Paper L- 101 L- 103 IC-101

G-101 M-102 H-101 S-101

Figure 100-2: Section 100 PFD

9 Stream Line Description Label Description Mass (1000 lb) Temp (°F) Pressure (psi) L-101 Dry Paper 10 75 15 L-102 Water 150 75 15 L-103 Paper-Water Slurry 160 75 15 Table 100-3: Stream Lines for Preparation

Equipment Description

Label Description Energy (1000 Btu/hr) Size (ft3) Cost (1000 $) IC-101 Inclined Conveyor (5hp) 13 - 11 H-101 Storage Hopper (5hp) 13 235 2 S-101 Shredder (10hp) 25 77 32 G-101 Conveyor (5hp) 13 - 11 M-102 Solid-Liquid Mixer (10hp) 25 880 42 Table 100-4: Equipment Details for Preparation

Hydrolysis is the process by which cellulose is converted into glucose and other simple sugars to enable the product to be fermented. This process can be performed through two methods. The first process, “acid hydrolysis”, requires a strong acid to break up the cellulose structure. The other process is “enzymatic hydrolysis” in which an enzyme and bacteria work together to break up the cellulose structure. Although acid hydrolysis provides approximately a 70 percent conversion from cellulose to glucose and enzymatic hydrolysis can provide up to 90 percent conversion, the acid hydrolysis process is more practical for this project [4].

In the hydrolysis process, sulfuric acid acting as the catalyst will be added to the paper slurry mixture at high temperatures in order to break the cellulose structure, resulting in the creation of more water as a bi-product. A liquid turbine will lower the pressure, while generating energy to power the heat exchangers in this process. Sodium hydroxide, a base, will then be added to remove excess sulfuric acid from the product. A rotary filter will be used to separate waste and solids that will result from this process. The result will be glucose and water, in preparation for the fermentation process.

Material Balance

After going through the paper handling and preparation process, the 160,000 pound slurry will be pumped with a ten horsepower slurry pump to a heat exchanger that will increase the temperature from approximately 75 oF to 400 oF. This requires 62,000 pounds of heated steam. 1600 pounds of sulfuric acid, at about 98 mass percent purity, will be heated to the same temperature and thus require 230 pounds of steam. Both components will enter the hydrolysis reactor at a heated temperature, which will enable the components to not require additional heating while in the hydrolysis reactor with a cooling jacket. The solution will be mixed together for less than one minute and will yield approximately 70 percent conversion of cellulose to glucose. The 161,600 pound mixture will then enter a heat exchanger where the temperature will be brought down to 212 °F. A pelton wheel, liquid turbine will lower the pressure from five atmospheres to one atmosphere and generate 16,000 kW of power which will then be recycled back to generate the power for the three heat exchangers in this section of the process. The mixture will then be sent to a mixer where 1,600 pounds of sodium hydroxide will be added to remove excess sulfuric acid. Neutralizing this solution will result in the formation of sodium sulfate and water. The chemical equation for this reaction is shown below.

H 2 SO4  2NaOH  Na2 SO4  2H 2O

Approximately 3,000 pounds of lignin and paper waste, as well as 2,300 pounds of sodium sulfate, are then removed through the rotary filter. The remaining 150,600 pounds of water and 5,600 pounds of glucose are ready for the fermentation process.

In Figure 200-1 below, the process flow diagram for hydrolysis is shown, identifying the mass and equipment required. The full Figure can be found in Appendix A as Figure A.200-1.

12 Sulfuric Acid 1600 lb

Heated to 400°F NaOH 1600 lb

150,000 lb H2O 5,600 lb Glucose

Heated to Expansion Chamber Mixer Rotary Filter Slurry Pump 400°F 160,000 lb slurry Hydrolysis Reactor

Solids/ Waste/ Salts 7,600 lb Figure 200-1: PFD for Hydrolysis with Masses Included

Energy Balance

Each stream and piece of equipment was then evaluated to determine the size required and amount of energy required throughout the process. Figure 200-2 shows the streamlines and equipment labels for the hydrolysis process. The full Figure can be found in Appendix A as Figure A.200-2.

Sulfuric Acid

L-204a L-204b

Base E-203 L-202a L-202b L-206 L-208

L-207 L-201 L-203 L-207 L-209 L-211 Paper Fermentation Handling

E-202 T-207 M-208 F-210 P-201 R-203 L-210

Figure 200-2: PFD with Streamlines and Equipment Labels

Table 200-1 below is directly related to Figure 200-2. Table 3 shows pertinent information for each stream line in the process. Mass is shown in pounds (lbs), temperature is represented in degrees Fahrenheit (oF), and pressure is represented in pounds per square inch (psi).

13 Stream Line Description Mass Label Description (lb) Temp (°F) Pressure (psi) L-201 Shredded Paper/ Water Mix 160,000 75 15 L-202a Steam 62,000 420 294 L-202b Condensate 62,000 420 294 L-203 Heated Paper/ Water Mix 160,000 400 74 L-204a Steam 227 420 294 L-204b Condensate 227 420 294 L-205 Sulfuric Acid 1,600 75 15 L-206 Heated Sulfuric Acid 1,600 400 74 L-207 Heated Hydrolized Mixture 161,600 400 74 L-208 Sodium Hydroxide 1,600 75 15 L-209 Neutral Mixture 163,200 400 74 L-210 Cooled Mixture 163,200 212 74 L-211 De-Pressurized Mixture 163,200 212 15 L-212 Solid/Waste Disposal 5,300 212 15 L-213 Glucose/Water Mixture 157,900 212 15 Table 200-1: Stream Lines for Hydrolysis Process

Equipment Sizing and Cost

Table 200-2 below shows the amount of energy required per day, the minimum size requirement, and the cost estimate for each piece of equipment. The energy for the rotary filter, mixer, and slurry pump were based on the amount of horsepower required for standard operation of the equipment. The amount of energy for both heat exchangers was calculated through changes in enthalpy, since steam will be used to heat the exchangers. It should be noted that the heat capacity used for water was 1.0 Btu per pound degrees Fahrenheit and the heat capacity used for paper was based off the cellulose value of 0.32 Btu/(lb oF) [5].

Equipment Description

Label Description Energy (1000 Btu/hr) Size (ft²) Cost (1000 $) E-202 Heat Exchanger for Slurry 2,083 2500 102 E-203 Heat Exchanger for Acid 7.7 10 3 F-210 Rotary Filter (5hp) 13 2,600 25 M-208 Mixer for Neutralysis (10hp) 25 2,700 40 P-201 Slurry Pump (10hp) 25 2,600 22.4 R-203 Hydrolysis Reactor 0 377 25.6 T-209 Pelton Wheel Turbine -6000 64 12 Table 200-2: Equipment Details for Hydrolysis Process

The total energy consumed for the hydrolysis process is 87,500 Btu/hr. This amount can be lowered through the use of recycle streams for greater efficiency. The volumes of each piece of equipment are rather large and can be reduced through a continuous process stream with less mass entering the system per given amount of time. The expected cost for this process is $200,000. Acid hydrolysis was chosen because of the low cost of acid and the confidence for which an undergraduate can perform the process. Enzymatic hydrolysis can be costly, must be performed under carefully monitored conditions, and requires an extensive biochemical understanding of enzymes. Other Considerations Enzymatic hydrolysis may provide a higher yield of glucose than acid hydrolysis and should be evaluated for further improvement. As technology advances, an enzymatic reaction may make sense for this process. At a large scale operation, the impact of the waste disposal should be considered, as well as determination for proper disposal, whether it be through incineration, landfill, or reusing the waste, primarily composed of lignin, as a heating source for other equipment in the process.

The solution coming from hydrolysis contains only about 3.6% glucose. However, in order to obtain maximum results from fermentation, a solution consisting of 8.6% glucose is desired. Evaporation is a means to increase the concentration of the glucose in the solution, and this section covers this process by examining the mass and energy balances as well as looking at economic considerations.

First a mass balance is done for the evaporator. Figure 300-1 shows the mass coming from hydrolysis to the evaporator and going out to the heat exchanger and the steam entering and leaving the evaporator. Figure 300-2 shows the heat exchanger which cools the solution before being sent to fermentation and shows the cooling water requirement for this process. Based on calculations shown in the results section, 60% of the water in the solution needs to be removed by the evaporator.

79,500 lb steam + 88,000 lb vapor

150,000 lb water 62,000 lb water 200: Evaporator Heat Exchanger Hydrolysis E-301 E-302 5,600 lb glucose 5,600 lb glucose

79,500 lb steam

Figure 300-1: Mass and Energy Balance – Evaporator

16 3,300 lb water

115°F

62,000 lb water 62,000 lb water Evaporator Heat Exchanger 400: E-302 E-301 5,600 lb glucose 5,600 lb glucose Fermentation

212°F 90°F

3,300 lb water

80°F Figure 300-2: Mass and Energy Balance – Heat Exchanger

Based on this mass balance the evaporator is sized according to the volumetric flow rate that it needs to handle. It is assumed that the evaporator will accommodate one hour’s worth of material at a time and that steam will be used to remove the excess water. Equation 300-1 [6] is used to determine the area needed for this evaporation process to take place:

A = q/(U*∆Tlm) (300-1)

Where A is the area in square feet, q is the energy requirement for the evaporator in 2 BTU/hr, U is the heat transfer coefficient in BTU/(hr ft °F) and ∆Tlm is the log mean square difference of the temperature change. The heat transfer coefficient used is based off heuristic values [7].

Next, the energy balance is examined for the evaporation process. The energy balance, seen in Figure 300-2 shows the energy required for the operation to take place and can give insight on whether a process is viable or where there are areas in the process that can be optimized for minimal energy consumption. The calculations for the energy balance are shown in the results section. Equations 300-2 and 300-3 are used to solve for the energy of the evaporator.

qadded  V (Hv)  F(Cp)T (300-2)

Fxf = Lxl + V (300-3)

Where λ is the latent heat of steam in BTU/lb Steam, Hv is the latent heat of water at 212 °F, Cp is the specific heat of water in BTU/(lb °F). F is the total liquid solution entering the evaporator, V is the vapor being removed from the evaporator, L is the liquid solution leaving the evaporator, xf is the fraction of glucose in F, and xl is the desired concentration leaving the evaporator

17 Based on the needs for fermentation the solution coming from the evaporator is at 212°F and needs to be cooled down to about 90°F. To do this a heat exchanger supplying cool water is used. Equation 300-4 is used to determine the amount of cooling water needed for the heat exchanger [8].

  mS (h1  h2 ) mw  (300-4) (h4  h3 )

Where mw is the mass flow rate of the water needed, ms is the mass flow rate of the solution, h1 and h2 are the enthalpies related to the change in temperature of the solution, and h3 and h4 are the enthalpies related to the change in temperature of the cooling water. Heuristics for heat exchangers were used to determine the values of h3 and h4. Once the mass flow rate of the water needed is determined, equation 300-5 is used to determine the heat requirement.

 (300-5) q  mw cpTlm

Figure 300-3 is a visual aide to help with the understanding of calculations being made on the evaporator. Figure 300-4 offers the same information as 300-3, except for the heat exchanger.

q + V added

F(x ) L(x ) f l 200: Evaporator Heat Exchanger Hydrolysis E-301 E-302

q added

Figure 300-3: Mass and Energy Balance Variables - Evaporator

m = Cooling Water w T = 115°F 4 h = 83 BTU/lb 4

m = mass flow soln m = mass flow soln s s Heat Exchanger 400: Evaporator Fermentation E-301 T = 212°F E-302 T = 90°F 1 2 h = 180.21 BTU/lb h = 58.05 BTU/lb 1 2 m = Cooling Water w T = 80 °F 3 h = 58.05 BTU/lb 3

Figure 300-4: Mass and Energy Balance Variables – Heat Exchanger

As stated in the introduction, the evaporator needs to remove approximately 60% of the water in the solution from hydrolysis. This is determined by using a modified version of Equation 300-3, dealing only with the liquid solution, as shown:

Fxf = Lxl 155,600(0.036) = L(0.083) L = 155,600(0.036/0.083) = 67,600 lb solution

19 L is the amount of solution in pounds that needs to go to fermentation, so to find the percentage of water to be removed, subtract 62,000 lbs (L minus the glucose in solution) from the 155,600 lbs of the original solution, then divide that value by the total solution coming:

Percent Water to be Removed = [(155,600-62,000)/155,600]*100 = 60%

Now, using the L and F values from above the missing value of V, the vapor product of evaporation, is determined.

F = L + V V = F – L = 155,600 – 67,600 = 88,000 lb vapor

With F, L and V known the energy input to the evaporator can be determined from Equations 2 and 3. The value of Hv is taken from available steam tables [8]. The enthalpy of the steam is used to determine the steam mass flow rate requirement.

lb BTU lb BTU BTU BTU q  88000 *977.6 155600 *1 *460  510R  93.4x106  3.9x106 day lb day lb * R day hr

BTU 93.6x106 day lb steam lb steam Steam   79500  3300 BTU day hr 1178 lb steam

Now that the energy requirement is known the area for the evaporator can be determined using Equation 1:

BTU day 93.6x106 * day 24hr A   116 ft 2 BTU (250  200)F 150 * hr * ft 2 * F ln250 / 200

Next the heat exchanger, E-302, requirements are determined. Using Equations 300-4 and 300-5, and the enthalpy values from water tables [8], the mass flow rate of the cooling water and the energy removed are found:

20 lb BTU 67600 180.21 58.05  day lb lb  water lb  water mw   330000  13750 BTU day hr 83  58.05 lb  lb  water BTU BTU BTU q  mCpT  330000 *1 *(90 115)F  8.25x106  344x103 day lb * F day hr

Table 300-1 shows a summary of the mass balance and Table 300-2 gives a summary of the energy balance. Each unit will be insulated so that heat lost to the atmosphere is assumed to be negligible.

Mass 1000 lb/day Destination F 155.6 To Evaporator L 67.6 From Evaporator V 88 From Evaporator Steam 3,300 lb/hr To Evaporator Water 13,750 lb/hr To Heat Exchanger Table 300-1: Mass Balance Summary

Heat 105 BTU/hr Unit

qadded 34 Evaporator

qremoved 3.4 Heat Exchanger Table 300-1: Energy Balance Summary

Once sizing and energy requirements were found cost of the equipment was considered next. First among these costs is the actual purchase price of the equipment. An evaporator capable of handling the quantity of solution has been difficult to find. Instead, multiple evaporators were priced that could handle the mass of the solution. This cost for the multiple evaporators came to about 1.8 million dollars.

Based on the number of evaporators needed and the energy requirement to run the evaporators it is recommended that the solution go to fermentation as it is without evaporation taking place. The reboiler in the distillation column will assume the energy cost of the evaporator, but will keep overall costs down by cutting out the cost of purchasing and installing an evaporator.

For now, more research needs to be done to optimize the evaporation process. All calculations done for the evaporator were for a single effect system. If a multiple effect evaporator is used, it should reduce the energy input to run the process significantly.

The solution coming into fermentation will be based around two scenarios. The first scenario will assume an evaporator is being used and the second will assume no evaporator. A brief introduction of fermentation will be given before the two scenarios are discussed.

A mass balance surrounding the fermentation vessel is done first and used as the basis for this section. Figure 400-1 illustrates the equipment used in the fermentation section. Insulated fermentation tanks are being used, so heat loss is considered to be negligible.

Yeast V-401

Mixer CO 2 V-402 V-403

300: Fermentation Filter 500: Evaporation R-401 F-401 Distillation

Figure 400-1: Fermentation Equipment

The sugar solution comes from Section 300 enters the fermentation tanks, R-401. Based on research done in the labs fermentation time per tank is around 12 hours [9]. With this in mind and assuming the process will run 24 hours a day, a minimum of three fermentation vessels will be needed for this process: one for the active fermentation and the other two for storage. When the sugar solution is in active fermentation vessel a portion of the warm solution will be sent to the mixer, V-402, where it will be mixed with the yeast stored in V-401. This mixing of the yeast activates the yeast and prepares it to be introduced into the fermentation vessel.

During the process of fermentation ethanol and carbon dioxide will be produced. The amounts are determined in part by Equation 400-1 which is the overall reaction for fermentation.

  C6H12O6 yeast 2(C2H5OH) 2(CO2) (400-1) Glu cos e Ethanol CarbonDioxide

23 Based on the stoichiometry of the chemical equation it can be seen that two times as much ethanol and carbon dioxide will be produced from the glucose coming into the system. The carbon dioxide produced will be collected and stored in V-403 and sold at current market value. Once fermentation is complete the yeast will be filtered, F-401, before being sent to distillation.

Table 400-1 gives all material constants used in Section 400.

Material Density, ρ (lb/ft3) Molecular Weight (lb/lb-mole) Glucose 96.1 180 Water 62.4 18 Ethanol 49.2 46 Carbon Dioxide 0.124 44 Table 400-2: Material Constants

Scenario One

This scenario details the expectations of the process with an evaporator. The primary benefit of having the evaporator is to get the concentration of the glucose in the solution entering the process at the ideal concentration of 8.3%. With this concentration it is believed that fermentation will take place the fastest for the Papernol process.

As stated in the 300 Section of this report with evaporation approximately 60% of the water from Section 200: Hydrolysis is removed. This significantly reduces the size of the fermentation vessels. The basis for the size of the fermentation vessels is determined by the pounds of solution coming into the fermentation and all components added to the solution such as yeast, based on the material constants in Table 400-1. Figure 400-2 illustrates the mass balance with the evaporator.

2,740 lbs CO 844 lb yeast 2

62,000 lb water Fermentation 62,000 lb water 62,000 lb water R-401 Filter 500: 5600 lb glucose 2,860 lb ethanol F-401 2,860 lb ethanol Distillation 844 lb yeast

Figure 400-2: Fermentation Mass Balance with Evaporator

Equation 400-2 is used to determine the volume of the fermentation tanks.

24 lb water lbethanol Vtotal   (400-2) (lb water / ft 3 ) (lbethanol / ft 3 )

The volume could have been determined by using glucose, but because of the difference in the densities of the two the volume turns out to be roughly the same. The calculated volume will be shown in the results section.

Scenario Two

This scenario details the expectations of the process without an evaporator. Without the evaporator the concentration of the glucose in the solution entering the process is the same as what exits from Section 200 which is 3.6%. The water from Hydrolysis is sent to fermentation. This significantly increases the size of the fermentation vessels versus scenario one. The basis for the size of the fermentation vessels is determined by the lbs of solution coming into the fermentation and all components added to the solution such as yeast, based on the material constants in Table 400-1. Figure 400-3 illustrates the mass balance without the evaporator.

2,740 lbs CO 844 lb yeast 2

150,000 lb water Fermentation 150,000 lb water 150,000 lb water R-401 Filter 500: 5600 lb glucose 2,860 lb ethanol F-401 2,860 lb ethanol Distillation 844 lb yeast

Figure 400-3: Fermentation Mass Balance without Evaporator

The volume of the fermentation vessels under these conditions will also use equation 400-2 listed above.

A note on the yeast requirement staying the same for both scenarios; the yeast needs are based solely on the glucose content in the solution, which stays the same for both processes. The yeast requirement is has been determined by previous research [2] shown in Table 400-2. With these values and the known amount of glucose entering the system, the yeast requirement for the process is determined by equation 400-3. Calculations for the yeast requirement can be seen in the results section.

Yeast Ratios 0.11 lb yeast/gal soln 0.72 lb glucose/gal water Table 400-3: Yeast Requirements

25 0.11lbyeast gallon solution Yeast  (lb glu cose) (400-3) 0.72lb glu cose gallon solution

The process of fermentation is dependent on the chemical Equation 400-1. This equation shows how much ethanol and carbon dioxide will be produced from glucose present in the solution. The numbers derived from Equation 400-1 will be the same for both scenarios.

  C6H12O6 yeast 2(C2H5OH) 2(CO2) Glu cos e Ethanol CarbonDioxide

5600lb glu cose  31lb  mole glu cose 180lb / lb  mole

31lbmole glu cose yeast62lbmoleethanol  62lbmoleCO2

 46lbethanol  62lbmoleethanol   2860lbethanol  lbmoleethanol 

 44lbCO   2  62lbmoleCO2    2740lbCO2  lbmoleCO2 

Now that the pounds of ethanol are known the volume can be calculated for both scenarios using Equation 400-2

62000lb water 2860lb ethanol V , S1    1050 ft 3 total (62.4lb water / ft 3 ) 49.2(lbethanol / ft 3 )

150000lb water 2860lbethanol V , S2    2460 ft 3 total (62.4lb water / ft 3 ) 49.2(lb ethanol / ft 3 )

A summary of the volumes is presented in Table 400-3. The conversion to gallons from cubic feet is 7.48 gal/ft3. As mentioned in the introduction of this section, a minimum of three tanks will be needed. The last two columns in Table 400-3 show the cubic feet and gallons per tank, respectively.

Volume ft3 gal ft3/tank gal/tank Scenario 1 1155 8700 385 2900 Scenario 2 2700 20300 900 6800 Table 400-4: Summary Fermentation Tank Volume

27 The yeast requirement shown by Equation 400-3 will be calculated next. It should be pointed out that the yeast does not play a part in the total volume of either scenario because of the packing nature of the yeast in solution. To compensate for the yeast and overflow a 10% correction factor is added in to the volumes in Table 400-3

0.11lbyeast gallon solution Yeast  (5600lb glu cose)  844lb yeast 0.72lb glu cose gallon solution

The main cost associated with fermentation is that of the fermentation tanks. Each tank is going to cost around $16,000 with the evaporator and about $29,000 without the evaporator. The extra $39,000 spent on the larger fermentation tanks without the evaporator is more cost effective than purchasing an evaporator that runs $1,800,000. The energy burden of processing that much water is shifted to the reboiler in the distillation column, but the column size does not have to be increased. Overall, it makes more sense to go without the evaporator unless one of two things happens:

1. The evaporator can be purchased at a significantly reduced price, or 2. Fermentation is adversely affected by the lower concentration of glucose in solution.

Laboratory tests will need to be performed to determine how fermentation will be affected by the various concentrations of the glucose in the solution.

There is room for optimization in this process. One item being considered is the recovery and reuse of the yeast from fermentation. A problem associated with recovering and reusing the yeast is making sure not to kill the yeast colonies in the process. Once again, laboratory experiments are required to test this procedure. Another idea for process optimization deals with scenario two. If the solution coming from hydrolysis isn’t cooled down before coming to fermentation, the solution could be rerouted to the reboiler to help compensate for the heat requirement there. This idea is currently being examined and will be reported if any significant breakthroughs are noted.

In the case of the Papernol process, distillation involves separating ethanol from water. This is known as binary distillation. It was decided that a continuous process would be the most beneficial and cost effective operation. Batch distillation is nearly always more expensive and time consuming. It was decided beforehand that a partial condenser and reboiler would work for the project. Vapor is needed coming out of the top of the column, and the water coming from the reboiler can be used for other purposes (such as basic heat transfer applications).

Two scenarios were looked at for the column. One scenario had an evaporator before the column and one did not. Both scenarios were discussed in Sections 300 and 400 above. One purpose of the evaporator was to lower the amount of water that would be coming into the column. The primary model discussed in this section will use the decision that no evaporation will be used. However, the other option with evaporation will be discussed along with the chosen design option.

The feed coming into the column from Section 400 contains 149,000 pounds of water and 2,862 pounds of ethanol. With the evaporator in place, there was less than half that amount of water. (Note: all units, unless otherwise specified, are on a per day basis.) This is an ethanol mole fraction of 0.0074. The distillation can only be carried out to the azeotrope of the mixture. The azeotrope in this case, given the standard atmospheric pressure that is being used and the lack of an additive, is 0.88. Since the ethanol feed coming into the column is so small, it was decided that having a bottoms composition of 0.0001 was a reasonable amount, considering the large volume of water being handled. Using this data, along with the Van Laar-Virial model data determined earlier and Dr. Cunningham’s DISTIL program [10], it was determined that the column would be 44 trays high with the feed coming in at tray 41. The scenario with the evaporator in place had 3 fewer columns.

All calculations will assume pound-moles (lbmols) are used unless otherwise noted.

Distillation Column Material Balance

Using the known inputs of the feed amount and concentration, and the desired output concentrations, it is possible to determine the amount of distillate, D, and bottoms product, B. This is done by a simple mass balance around the column. It was determined from these calculations that there are about 70 moles of 0.88 concentration distillate product and 8300 moles of 0.0001 concentration bottoms product.

Using the desired reflux ratio of 42, it can easily be determined what is going on inside the column as far as fluid flow goes. The reflux value is the ratio of the liquid flow inside the column, L, over the distillate amount. This means that L is 2928.7 moles. The vapor flow inside the column, V, is the liquid flow amount plus the distillate amount, or 2998.4 moles. Interestingly enough, between the two scenarios, there is practically no difference in the value of V and L. This is because the total distillate amount is approximately the same and the reflux ratio is the same. Using the solute free analysis, it can be determined what the solute free vapor flow, V’, and the solute free liquid flow, L’, are inside the column. Involved in both of these value calculations is the fraction of the feed that vaporizes upon entering the column, f. This value is determined by Equation 500-1.

CPf *(TBP TF ) f   (500-1) F

CPF is the molecular average of the specific heat, TBP is the temperature of the boiling o point, TF is the feed temperature (in this case, 189 F), and F is the molar average latent heat of vaporization for the feed. Using these values, it was determined that f is equal to -0.0164. This means that most of the feed is vaporized upon entering the column, but a small amount remains liquid. Using this value, and Equations 500-2 and 500-3, the solute free vapor and liquid flows can be determined.

L' L  (1 f )* F (500-2)

V ' V  f * F (500-3)

V’ was determined to be 3135.5 moles and L’ was determined to be 11426.8 moles. These values are used in the next step: energy balances on the reboiler and condenser.

Distillation Column Energy Balances

The purpose of the energy balances on the column is to determine the energy requirement to vaporize V in the reboiler and condense V in the condenser. The easiest way to conduct these calculations is to assume that V in the reboiler is nearly pure water (which it is) and that around the condenser it is nearly pure ethanol. The heat of vaporization for pure water is 17,392 Btu/lbmol and for ethanol is 16,319 Btu/lbmol. V is approximately

31 3000 mols which means the energy requirement of the condenser is 48,957,000 Btu and the energy requirement for the reboiler is 52,176,000 Btu.

These values for both the reboiler and the condenser would work for any column that has approximately 70 moles of 0.88 concentration solution coming out as distillate. Having the evaporator in Section 400 will not affect the values of heat input required by the reboiler or the condenser.

Sizing the Column

The next issue to design was the size of the distillation column. The operations book used in class [11] has a very useful example of exactly how to do this. This first step is to determine the density and viscosity of the liquid solute free flow within the column. This is done by molecular averages. The density is approximately 64 lb/ft3 and the viscosity is about 61.8 dyne/cm. The molecular weight is 18.02 lb/lbmol and the total mass of L’ is about 206,000 lbs. Doing simple unit cancellations gave the values for the vapor solute free flow. The molecular weight is about 18.03 lb/lbmol. The total mass is just under 62,400 lbs and the density is about 0.0368 lb/ft3. These values can be used in Equation 500-4 which is used to determine the value for the flooding constant. For both scenarios, with the evaporator and without, this value is about the same, which ultimately means the value for the size of the column is the same.

L  0.5 ( V ) (500-4) V  L

Using this value on graph 21.26 in the Unit Operations of Chemical Engineering text [11] will give a value for Kv, which is a flooding constant used to determine percent flooding. Papernol assumes a tray spacing of 12 inches. Going to the proper line on the graph will allow you to solve for the maximum allowable vapor flow rate in the column. In the case of Papernol, it is 11 ft/s. The next step is to solve for the bubbling area of the column. This is done by taking the vapor volumetric flow rate over the maximum allowable flow rate. The vapor volumetric flow rate is solved by Equation 500-5.

D *(R 1)* MW V  (500-5) 

Where MW is the molecular weight, R is the reflux ratio, D is the distillate flow, and  is the density. The flow rate is about 17 ft3/s. This gives a bubbling area of 1.55 ft2. For Papernol, it is assumed that the bubbling area is seven tenths of the area of the tray meaning the column has to have a cross-sectional area of 2.21 ft2. Solving this for the diameter means the column is about 1.69 ft in diameter.

The height of the column is the number of trays multiplied by the distance between each tray, 1 ft. Therefore, the column is 44 ft high and about 1.69 ft in diameter. Most likely, this column will have to be split up into multiple columns and the distillation will be done continuously in stages.

32 Molecular Sieve Description

A molecular sieve is an adsorber that adsorbs certain molecules while allowing others go by untouched. In the case of Papernol, the sieves adsorb water but leave the ethanol in the stream. The first step in designing these sieves is to do the material balance around them.

Molecular Sieve Material Balance and Sizing

It is first necessary to determine what material is going to be used as the adsorbent. According to several sources, material 3A is best suited to adsorbing water, but not ethanol [12]. 3A has a density of about 46 lb/ft3 and a water adsorption capacity of 21.5% of its weight. It is assumed for this project that the residence time in each adsorber is going to be about one hour. There are two adsorbers for the process; one is running while the other is being regenerated. (Note: Since the calculations of an adsorber contain excess adsorbent anyway, the accuracy of the calculations is not entirely necessary. This means a distillate value of 70 moles is going to be used.) Since the amount of water having to be adsorbed is about 6.3 lbs per hour and the adsorber can take up to 21.5% of its weight, there needs to be at least 29 lbs of adsorbent material. Using a cautious value, it is assumed that 60 lbs of 3A material is going to be used. With a density of 46 lb/ft3, this means the sieve needs to have a volume of 1.3 ft3. The simple sieve that can be considered is one that is 1 ft long and has a cross-sectional area of 1.3 ft2.

Molecular Sieve Energy Balance

The primary energy requirements come from the regeneration of the sieves. The regeneration involves removing the water that was adsorbed from the input stream. The way this is done is by blowing hot dry air through the sieve and evaporating the water into the air. It is assumed that the air coming in is at 225 oF and coming out at 212 oF. Also assumed is the volumetric flow rate of 3 ft3/min.

The energy required for heating the air is determined using Equation 500-6.

Q  CP *T * M (500-6)

The air is being heated from 77 oF to 225 oF for 40 minutes to regenerate, and then from 77 to 212 oF for cool-down. Cool-down is done to get the adsorbent to the temperature of the vapor coming in so that no desired material is lost. In total, it will take about 425,000 Btu/hr to heat the air required.

Another source of energy loss is the condenser. The nearly pure ethanol that comes out of the sieves comes out as a vapor. To cool the mixture from 212 to 77 oF is approximately 9,300 Btu/hr. This was determined using Equation 500-6. After all of this, there is 2,838 lbs of ethanol that has been condensed and is ready for use or sale.

33 Cost of Equipment

CAPCOST [13] is the program that was used for this analysis. This program was used to get a bare bones idea of how much everything would cost. This means that the most basic materials were used and the cost analysis was not as detailed as it could have been. According to the program, for a column that is 44 feet high, the purchased equipment cost is $100,000 while the bare module cost is $600,000. The sieves, because they are so much smaller, are much cheaper with a purchased equipment cost of $1,300 and a bare module cost of $10,000.

The Distillation column can output 70 moles of 88% ethanol per day. The column required to accomplish this is 44 ft tall with 44 trays and the feed entering the column at tray 41.

The molecular sieves will output 65 moles of 99% ethanol per day. Each of the two sieves will have to be 1.3 ft3. The adsorbent will require regeneration every hour. During this time, the other molecular sieve will be used so that he process can be continuous.

One of the most important parts of a plant design is the utilities. The Papernol plant consists of three main utilities: low pressure steam (produced from natural gas), electricity, and water. For the purposes of this plant, there will be a reboiler distributing the steam to the needed sections and a cooling tower distributing the water.

Figure 600-1: Block Diagram of the Reboiler with Each Section’s Inlet and Outlet Stream Lines

Figure 600-1 shows each section’s steam requirement with the production and distribution of steam. The left side of the reboiler depicts the stream lines that are being recycled back to Section 600 after leaving Sections 200 and 300. These are the only two sections that will produce steam within the process. The right side of the reboiler is the stream lines for each section that has a demand for steam. These sections are 200, 300, and 500.

Figure 600-2: Block Diagram of the Cooling Tower for Supply and Recycle of Water

36 Figure 600-2 shows the pipelines labeled by section that will need water from the cooling tower. It also shows that for each section, the water coming into the system will also for the most part be leaving the system as well.

Recycle streams for the water and steam circulating throughout the system play an important role in the cost of producing the 500 gallons of ethanol per day. Without a recycle stream for water, thousands of dollars is being spent each day just to maintain the required amounts of water for the production of the ethanol. These differences will be further examined in the results.

Another costly piece of equipment is the evaporator. With the evaporator, the concentration of the glucose is more ideal for the fermentation, however, it significantly decreases the amount of water being recycled back into the system through the distillation column. This recycle stream would potentially be supplying 99% of the water into the system. Also, the evaporator will require water and steam, leading to higher production costs for the plant. The cost to produce one gallon of ethanol will be examined with the recycle line and evaporator and without to see which is less costly, more beneficial, and if these two sources are a necessity to the plants operation.

In the table below are the prices for electricity, cooling water, and low pressure steam. These prices are the price of the utilities as of 2008 in Chattanooga [14-16].

Utility Description Cost $/common unit Cost $/Papernol Unit Steam Low Pressure ( 87 psi, 363 ºF) $6.40/1000 kg $0.0058/lb Cooling Water Cooling Tower Water $.310/GJ $0.0059/lb Fuel Natural gas $2.50/GJ Electricity $0.08/kWh $0.000023/BTU Table 600-1: Utility Prices for the Papernol System

Utilities With Evaporator Section Electricity (1000 BTU) LPS (1000 lb) Water (1000 lb) 100 89 0 160 Cost ($) 2 - 944 200 63 62.23 0 $ 1 361 - 300 0 79.5 79.2 Cost ($) - 461 467 400 0 0 0 $ - - - 500 10,200 16.3 98.8 Cost ($) 235 95 547 Total Cost ($) 238 917 1958 Total System Cost ($) 3113 Table 600-2: Table of Utility Prices with Evaporator Process

Using the values in Table 600-2, the total price for operating the system per day with the evaporator process is roughly $3100. This means that the cost to produce one gallon of ethanol using the evaporator process would be estimated at $6.25.

38 Utilities Without Evaporator Section Electricity (1000 BTU) LPS (1000 lb) Water (lb) 100 89 0 160 Cost ($) 2 - 944 200 63 62.23 0 $ 1 361 - 300 0 0 0 Cost ($) - - - 400 0 0 0 $ - - - 500 10,200 16.3 98.8 Cost ($) 235 95 547 Total Cost ($) 238 456 1494 Total System Cost ($) 2188

Table 600-3: Utility Prices for the System without an Evaporator

From Table 600-3 it can be seen that eliminating the evaporation process from the Papernol system reduces the total price of utilities per day by $2200 This will make the price to produce one gallon of ethanol go from $6.25 to $4.40.

The next step is to examine the effects a recycle stream will have on the price of ethanol with and without an evaporator.

Recycle Stream With Evaporator Section In (lb) Out (lb) 100 150,000 0 200 0 61,773 300 79,200 79,200 400 0 0 500 0 62,000 Excess Water/day -26,227 Table 600-4: Excess Water in Recycle Stream with the Evaporator Process

While the evaporator process does concentrate the sugar solution going into the fermentation subsystem, it also decreases the amount of water that is being circulated throughout the system. According to Table 600-4, roughly 26,000 lbs of water will be needed from utilities per day to satisfy the water requirements. This will cost an extra $150 dollars a day.

Recycle Stream Without Evaporator Section In (lb) Out (lb) 100 150,000 0 200 0 61,773 300 0 0 400 390 0 500 0 149,000 Excess Water/day 60,773 Table 600-5: Excess Water in Recycle Stream without the Evaporator Process

By eliminating the evaporator from the Papernol process, there is an excess of 61,000 lbs of water per day. Once the recycle water is in circulation within the system, the need for water from utilities is obsolete, reducing the costs significantly for the system.

Utilities With Recycle and Evaporator Section Electricity (1000 BTU) LPS (1000 lb) Water (1000 lb) 100 89 0 160 Cost ($) 2 - 0 200 63 62.23 0 $ 1 361 - 300 0 79.5 79.2 Cost ($) - 461 0 400 0 0 0 $ - - - 500 10,200 16.3 98.8 Cost ($) 235 95 0 Total Cost ($) 238 917 +150 Total System Cost ($) 1305 Table 600-6: Utility Costs with a Water Recycle Stream and an Evaporator

Highlighted in yellow is the cost of the production of the 26,000 lbs of water still needed in the process per day. With a recycle stream and evaporation process, the price of ethanol drops from $6.25 to $2.60. This is a significant decrease in the price of production for the ethanol; however, it is still too high to make the Papernol system feasible and profitable.

Utilities With Recycle and Without Evaporator

40 Section Electricity (1000 BTU) LPS (1000 lb) Water (1000 lb) 100 89 0 160 Cost ($) 2 - 0 200 63 62.23 0 $ 1 361 - 300 0 0 0 Cost ($) - - - 400 0 0 0 $ - - - 500 10,200 16.3 98.8 Cost ($) 235 95 0 Total Cost ($) 238 456 0 Total System Cost ($) 694 Table 600-6: Utility Costs for the System with a Recycle Stream and without the Evaporator Process

Table 600-6 shows the total cost for the system per day with the recycle stream and without the evaporator. With these parameters in combination, the price to produce one gallon of ethanol drops from $4.40 to about $1.40.

In terms of meeting the goal initially set of producing one gallon of ethanol for $1.50, the process can realistically produce ethanol for an estimated $1.40 a gallon. With a water recycle stream and without the evaporator process, the price to produce one gallon of ethanol is less than our initial expectations. The possibility for a steam recycle stream could effectively lower this production price even further, potentially beating Kelly Tiller’s expected production cost of $1.25 per gallon.

42 Conclusions

Each section in the Papernol process came to individual conclusions. These included decisions about material inlets and outlets, material cost, energy balances, energy cost, equipment sizing, and equipment cost. These decisions were important and not necessarily independent of other sections. Some conclusions were made about the economics and the environmental aspects of the process.

Section 100

A basis of 10,000 lbs of paper per day was selected based on the daily usage of the current Orange Grove Recycling Center. This value set the general scale of the project. Later, this amount can be scaled up or down to meet other needs. The amount of water was set at a ratio of 15:1, and thus 150,000 lbs of water was added to a slurry mixture. This mixture could then be sent to Section 200.

Preparation of the paper requires many unique pieces of equipment. A 77 ft3 shredder is needed to break the paper into digestible pieces. Two conveyors send the paper from the initial paper storage to the shredder and from the shredder to the 880 ft3 mixing tank.

The energy required for this section of the process is minimal. Only electrical power to the previously mentioned equipment is required. A total amount of 217 kW is required.

The total cost for this section is about $98,000.

Section 200

Using the 160,000 lbs of slurry from Section 100 as the feedstock, 1,600 lbs of sulfuric acid must be added in the hydrolysis process. The two streams must be heated before being mixed. This heating is accomplished with fresh steam at 420 oF coming into the hydrolysis process and the excess heat will be recycled within the system. This accomplishes two goals: less hot steam and therefore energy is needed for the section and the product stream can be cooled and depressurized. The strong base sodium hydroxide is added to the stream to neutralize the acid. A pelton wheel turbine produces mechanical energy that can further be recycled in the system. The last step before going to Section 300 is the rotary filter which removes lignens and clay imbedded in the paper along with other solids including salts from the neutralysis.

At least two heat exchangers with feedback controls will be required to heat the two input streams up to 400 oF. The other equipment has already been mentioned and includes the hydrolysis mixer, pelton wheel turbine, neutralysis mixer, and rotary filter. A pump will also be needed to maintain pressure in the hydrolysis vessel.

The biggest energy burden comes from the energy necessary to heat the mixture. This can partially be aided by recycling heat whenever possible. Electrical energy will be required

43 for mixing vessels, rotary filter, and pump. However, the pelton wheel turbine will generate energy, so this can offset some of the energy cost.

Section 300

The feedstock for this section is the 3.6% glucose-water mixture. If evaporation is not used, this stream goes straight through to Section 400. This would be applicable if several conditions are verified. These could be things such as excess water not affecting the fermentation yield and the water being able to be removed later in the process through distillation. If evaporation does occur, 62,000 lbs of water is removed from the process with 17,000 lbs of cold water and steam in a heat exchanger. The final mixture would then be an 8.3% glucose-water mixture.

The only equipment in this section is an evaporator costing approximately 1.8 million dollars.

The total cost for this section with the evaporator is 1.8 million dollars contrasted with no cost without the evaporator.

Section 400

It was decided that the un-evaporated solution of 3.6% glucose will be the feedstock to the fermentation section. Using three fermentation vessels on a semi-batch rotation, yeast will be added to the solution in each tank. After 12 hours of fermentation, the converted ethanol-water solution will be sent to Section 400. The fermentation process also releases CO2 gas that will be contained.

The equipment for this section consists of the three fermentation vessels costing about $29,000 each.

Section 500

Once the cellulose from the paper has been converted to glucose in Section 200 and subsequently to ethanol in Section 400, it is ready to be separated. This is to ensure that the ethanol produced is of the highest quality at about 99% purity. First, the solution is sent to a distillation column. The distillation column produces a distillate of 88% ethanol and a bottoms of .01% ethanol. The distillate is sent to staged molecular sieves where water is further removed from the system. Eventually, 3000 lbs of 99% ethanol will leave the molecular sieves. The water removed can be recycled. The .01% bottoms can be further reboiled to attempt to conserve all of the ethanol. The bottoms can also be recycled back to the system where water is needed.

44 The main piece of equipment is a 44 tray distillation column with a partial reboiler and partial condenser. This is estimated to cost $100,000. The other equipment is two molecular sieves that cost about $1,300 each.

The distillation column requires more energy than any other part of the process. This amount is increased if the reboiler takes on the burden of removing the water in the solution when the evaporation step was skipped. There are other heat exchangers needed to keep the distillate at the correct temperature before going to a molecular sieve.

Section 600

The final section of the Papernol process is utilities. The burden of the utilities section is to supply each of the other sections with the required energy. The energy can be supplied with electrical energy, natural gas, or water. The water can be used as either cooling water or low pressure steam if used with natural gas.

The utility costs will be for the entire system. The cost of electricity required is $238. The maximum cost for the use of natural gas to produce steam for the system per day will be $917. The maximum cost for water for the process will be $1,958.

The maximum total cost of this section is $3,113 per day. However, if there is a recycle stream and no evaporation process, this total cost decreases to $694 per day.

Economic Feasibility

The total cost for the Papernol equipment with the evaporator process is calculated to be $2.3 million dollars. Without the evaporation process, the total cost for the Papernol equipment is $518,000.

After ten years with an assumed interest rate of 10%, the annual worth on the invested $3,000,000 to build the Papernol plant is roughly $600,000. The future worth for this project at this interest rate will be almost $12.2 million dollars.

Depending on the presence of the evaporation process, the price to produce one gallon of ethanol varies from $6.25 to $1.40.

Environmental Feasibility

In the undertaking of the Papernol process, lots of energy will be consumed. When energy is used or given off, it makes an impact on both the surrounding and global environment. There will be 2,740 lbs of CO2 being produced each day from the fermentation subsystem. Without proper ventilation, this amount of CO2 can negatively

45 impact on the surrounding environment. Proper containment of this CO2 is mandatory for the Papernol process.

46 Recommendations

After designing the Papernol system it is found that the system is more efficient and less costly with a water recycle stream and without and evaporation process. This process will produce an excess amount of water that can be put to various uses within the system such as the production of steam. One improvement for future examination would be the use of a steam recycle stream in the process. This could further reduce the utility costs for the system.

47 References

1. Energy Information Administration. 2. Lara Amini, Josh Oliver, Steve Wild, Papernol: Cellulosic Ethanol from Waste Paper Feedstock, (2008), UT – Chattanooga, Chemical Engineering Dept 3. “Density of Paper and Paperboard.” 4. Fagan, R. D., “The Acid Hydrolysis of Refuse,” M.E. Thesis, Thayer School of Engineering, Hanover, N.H., July 1969. 5. linkhub.elsevier.com/retrieve/pii/0040603194020026 or Google: “cellulose heat capacity”; Actual value provided 1338 J/(kg*K) 6. Christie Geankoplis, Transport Processes and Unit Operations, (1993) 3rd ed., p. 496, Prentice Hall, Englewood Cliffs, NJ 7. Turton, Richard et al, Analysis, Synthesis, and Design of Chemical Processes, (2003) 2nd Ed, p .345, Prentice Hall, Upper Saddle River, New Jersey 8. Boles, Cengal, Thermodynamics: An Engineering Approach, (2006) 5th Ed, p. 940-944, McGraw Hill, New York, NY 9. Jim Henry, email correspondence, April 2008, UT – Chattanooga, Chemical Engineering Dept 10. Distil Program 11. Unit Operations 12. Sieves 13. CapCost 14. http://www.amwater.com / 15. http://www.chattanoogagas.com/ 16. http://www01.epb.net/

48 Appendices

Figure A.200-1: PFD for Hydrolysis with Masses Included

Figure A.200-2: PFD with Streamlines and Equipment Labels