SUGAR PRODUCTION.pdf

Kingdom of Saudi Arabia King Saud University College of Engineering Department of Chemical Engineering

SUGAR PRODUCTION USING THE CARBONATION PROCESS INCORPORATING WATER RECYCLING AND TREATMENT UNIT

Submitted by

SULTAN AL AHMARI (428102729) MUBARAK AL QAHTANI (428102448) ALI AL GHAMDI (428102690) ZIYAD AL SAKRAN (428102060)

Supervised by

Prof. MOURAD BOUMAZA Dr. MOHAMED KAMEL HADJ-KALI

1433 – 1434 H (2012 – 2013 G)

ACKNOWLEDGEMENT First of all, we would like to show our great thanks to our supervisors in this project Prof. Mourad and Dr. Mohamed Kamel for their help during all this semester to gain a good project. Also, we want to thank the entire employees of the Chemical Engineering Department for their efforts during the last 5 years and we will not forget any moment with anyone. Finally, we want to thank our parents and family members who supported us during this semester.

I

ACKNOWLEDGEMENT First of all, we would like to show our great thanks to our supervisors in this project Prof. Mourad and Dr. Mohamed Kamel for their help during all this semester to gain a good project. Also, we want to thank the entire employees of the Chemical Engineering Department for their efforts during the last 5 years and we will not forget any moment with anyone. Finally, we want to thank our parents and family members who supported us during this semester.

I

EXECUTIVE SUMMARY

The objective of this project is to carry out a design of plant that produces 300 ton/day of sugar from sugar beet. The carbonation process using sugar-beet has several benefits and advantages in comparison to the one using sugar cane. In addition to the main product (white sugar), the sugar-beet factories produce molasses and pulp which are used to prepare feed for cows and sheep. The plant consists of 15 main units. Five units have been fully designed, namely: 

The Continuous Continuous Stirred Tank Reactor (CSTR) (CSTR) ,



The Rotary Drum Filter,



The Multiple Effect Evaporator,



The Natural Draft Cooling Cooling Tower



The Rotary Drum Dryer

The remaining units have been sized. As the sugar process requires an large quantity of water, we have also studied the water treatment and recycling part to minimize the water consumption. A hazard and operability (HAZOP) analysis both with the environmental and safety have also been covered in this project. The economical study has shown that this project requires a capital cost of $98 million and about 3 years payback period to retain the total capital investment.

II

TABLE OF CONTENTS

Chapter 1. Introduction

3

.............................................................

1.1 Overview……………… ...........................................................................3 1.2 History of Beet-Sugar.. ...........................................................................5 1.3 By – Products of The Beet-Sugar Factory ..............................................6 1.3.1 Pulp…… ...........................................................................................6 1.3.2 Carbonation-Lime Residue ...............................................................6 1.3.3 Molasses…. ......................................................................................6 1.4 Sucrose Properties.. ...............................................................................7 1.4.1 Sucrose Formation………. ...............................................................7 1.4.2 Sucrose Properties.. .........................................................................7 1.4.3 Sucrochemistry… .............................................................................9 Chapter 2. Process Description

12

.................................................

2.1 Block Diagram ......................................................................................12 2.2 Flow sheet of Process.. ........................................................................13 2.3 Description of Flow sheet .....................................................................14 Chapter 3. Material Balance

15

......................................................

3.1 The General Equation of Mass Balance. ..............................................15 3.2 Mass Balance of Units .. .......................................................................16 Chapter 4. Energy Balance

21

.......................................................

4.1 Heat Capacities…….. ........................................................................21 Chapter 5. Equipment Design

..................................................

29

5.1 Reactor Design (Unit – 4) ……. ............................................................29 5.2 Filter Design (Unit – 7)… ......................................................................32 5.3 Evaporator Design (Unit – 8)… .............................................................35 5.4 Dryer Design (Unit – 13) . .....................................................................40 5.5 Cooling Tower Design (Unit – 14). ........................................................42 Chapter 6. Sizing Equipment

45

.....................................................

6.1 Sizing of Mixer (Unit -1). ....................................................................45 6.2 Sizing of Diffuser (Unit -2). ..................................................................46 III

6.3 Sizing of Heater (Unit-3). ....................................................................47 6.4 Sizing of Lime Unit (Unit - 4). ..............................................................48 6.5 Sizing of Clarifier (Unit - 5). .................................................................48 6.6 Sizing of Heater (Unit - 6). ...................................................................49 6.7 Sizing of Sulfitation (Unit - 9). .............................................................50 6.8

Sizing of Crystallizer (Unit - 10). ........................................................50

6.9 Sizing of Cooler (Unit - 11). .................................................................50 6.10 Sizing of Centrifuge (Unit -12). ..........................................................51 6.10 Sizing of Storage (Unit - 15). .............................................................51 Chapter 7. Water Recycling and Treatment

...................................

53

7.1

Overview. ...........................................................................................53

7.2

Washing water recycling. ..................................................................54

7.3

Diffusion Tower water recycling. .......................................................55

7.4

Evaporator Water Recycling To Diffusion Tower. .............................55

7.5

Cooling Water Recycling From Cooling Tower. .................................56

7.6

Water treatment . ...............................................................................57

Chapter 8. HAZOP and Environment

...........................................

58

8.1 HAZOP……. .........................................................................................58 8.1.1 Reactor ……… ...................................................................................59 8.1.2 Cooling Tower ……… .........................................................................60 8.1.3 Evaporator ……… ..............................................................................61 8.1.4 Dryer ……….. .....................................................................................62 8.2 Environmental.. .....................................................................................63 8.2.1 Pollution Control Strategies ............................................................63 8.2.2 Emissions to Air.. ............................................................................64 8.2.3 Safety… ..........................................................................................65 Chapter 9. Control Loop

67

..........................................................

9.1

Overview: ......................................................................................67

9.2

Diffusion Tower Control .................................................................68

9.3

Mixer Control .................................................................................68

9.4

Heater(1) Control ...........................................................................69

9.5

Liming Control ...............................................................................69

9.6

Clarifier Control .............................................................................70 IV

9.7

Heater(2) Control ...........................................................................70

9.8

Filter Control ..................................................................................70

9.9

Evaporator Control ........................................................................71

9.10

Sulfitation Control ........................................................................71

9.11

Crystallizer Control ......................................................................71

9.11

Cooler Control .............................................................................72

9.12

Centrifuge Control .......................................................................72

9.13

Dryer Control ...............................................................................72

9.14

Cooling Tower Control .................................................................73

Chapter 10. Cost Analysis

75

........................................................

10.1 Overview… .........................................................................................75 10.2 Methods for estimating capital investment ..........................................75 10.3 Equipment cost purchased in 2007..................................................... 75 10.4 Equipments In KSA at 2012 ................................................................76 10.5 Total Capital Investment (TCI)…............................. …………….……. 81 10.6 Total Production Cost (TPC)….. ........................................................83 10.7 Profitability Analysis ............................................................................86 10.8 Payback Period (PBP)…………………………………………………… 86 Chapter 11. Conclusion REFERENCES APPENDICES

88

...........................................................

89

.......................................................................

91

........................................................................

V

LIST OF TABLES Table 1.1 Annual Consumption of Sugar in The World ……………… ..........4 Table 1.2 Sucrose Properties .. ....................................................................8 Table 3.1 The Materials of Process………. . ...............................................15 Table 5.1 Reactor Design ……. ..................................................................32 Table 5.2 Filter Design … ...........................................................................34 Table 5.3 Evaporator Design… ..................................................................39 Table 5.4 Dryer Design …. . .......................................................................42 Table 5.5 Cooling Tower Design… .............................................................44 Table 6.1 Sizing of Mixer. ..........................................................................45 Table 6.2 Sizing of Diffuser ... .....................................................................46 Table 6.3 Sizing of Heater. .........................................................................47 Table 6.4 Sizing of Lime Unit . ....................................................................48 Table 6.5 Sizing of Clarifier . .......................................................................48 Table 6.6 Sizing of Heater. ........................................................................49 Table 6.7 Sizing of Cooler .. ......................................................................51 Table 6.8 Sizing of Centrifuge. ..................................................................51 Table 7.1 Mass Balance of Washing Water Recycling. .............................53 Table 7.2 Mass Balance of Diffusion Tower water recycling......................55 Table 7.3 Mass Balance Evaporator Water Recycling..............................55 Table 8.1 HAZOP of Reactor ……. .............................................................59 Table 8.2 HAZOP of Cooling Tower ...........................................................60 Table 8.3 HAZOP of Evaporator .................................................................61 Table 8.4 HAZOP of Dryer .........................................................................62 Table 8.5 MSDS of Sugar Factory .. ...........................................................65 Table 9.1 Control Overview. ......................................................................74 Table 10.1 Total Capital Investment… .......................................................82 Table 10.2 Raw Material………………….. ..................................................83 Table 10.3 Total Production Cost.. .............................................................84 Table 10.4 Calculations of Total Production Cost .......................................85 Table 10.5 Summary for Economics Calculation ……….…………….…….87

VI

LIST OF FIGURES Figure 1.1 Beet and Cane Distribution in The World ……………… ..............4 Figure 1.2 The Relationship Between Solubility and Temperature .. ............8 Figure 2.1 Block Diagram of Process………...............................................12 Figure 2.2 Flow Sheet of Process ……. ......................................................13 Figure 5.1 Reactor Design … .....................................................................29 Figure 5.2 Rotary Drum Filter … .................................................................33 Figure 5.3 Multiple Effect Evaporator …. . ..................................................35 Figure 5.4 Rotary Drum Dryer ..… ..............................................................40 Figure 5.5 Natural Draft Cooling Tower. .....................................................42 Figure 7.1 Diagram of Water Recycling and Treatment .............................53 Figure 7.2 Diagram of Washing Water Recycling. ......................................54 Figure 7.3 Diagram of Diffusion Tower Water Recycling . ..........................55 Figure 7.4 Diagram of Evaporator Water Recycling. ..................................55 Figure 7.5 Diagram of Cooling Water Recycling From Cooling Tower........56 Figure 9.1 Control Flow Sheet .. .................................................................67 Figure 9.2 Diffusion Tower Control . ...........................................................68 Figure 9.3 Mixer Control ... .........................................................................68 Figure 9.4 Heater1 Control . .......................................................................69 Figure 9.5 Liming Control. ..........................................................................69 Figure 9.6 Clarifier Control. .........................................................................70 Figure 9.7 Heater2 Control. ........................................................................70 Figure 9.8 Filter Control. .............................................................................70 Figure 9.9 Evaporator Control. ...................................................................71 Figure 9.10 Sulfitation Control. ...................................................................71 Figure 9.11 Crysallizer Control. ..................................................................71 Figure 9.12 Cooler Control. ........................................................................72 Figure 9.13 Cenrifuge Control. ...................................................................72 Figure 9.14 Dryer Control. ..........................................................................72 Figure 9.15 Cooling Tower Control. ............................................................73

VII

NOMENACLATURE

P

Purity

T

Temperature

DS

Dry substance

m

Mass Flow rate

CP

Heat Capacity



Density



Latent Heat

H

Enthalpy

Q

Amount of heat

U

Overall heat transfer coefficient

∆Tlm

Log mean temperature difference

V

Volume

ha

Outlet Humidity of air

hb

Inlet humidity of air

X

Conversion of reaction

Xa

Initial moisture

Xb

Final moisture

D

Diameter

L

Length

G

Mass velocity of the dry air

t

Time

C

Concentration

F

Flow rate in reactor

ᵋ

Percentage of mole change

f

Submergence fraction

α0

Cake resistance

µ

Viscosity

∆P

Pressure drop

s

Compressibility coefficient

VIII

Chapter 1. Introduction 1.1 Overview "Beet and cane are almost similar in sugar content (beet typically contains 18% and cane about 15%), but are dissimilar in the amount of non sugars (beet juice contains about 2.5% and cane juice about 5%) and fiber (beet contains about 5% and cane about 10%). The composition differences require different methods to produce sugar from beet or cane. The differences in farming, composition, and processing of these crops are sufficient to justify two separate industries: 

Beet-sugar industry



Cane-sugar industry

During the last century, both industries have grown considerably. World sugar production has increased from approximately 10 million tons in 1900 to about 160 million tons in 2012. There are over 500 beet-sugar factories in the world. About 40% of the world’s sugar production is from beet, and 60% is from cane. The climates of

most sugar-producing countries are suitable for growing either beet (in moderately cold areas) or cane (in tropical areas). In only a few countries (United States, Iran, Spain, Egypt, and Pakistan), the growing conditions are suitable for both crops. Sugar from sugarbeet is produced in about 50 countries worldwide, in North America(United States and Canada), South America (Chile), Asian, North Africa (Morocco and Egypt)countries, and most of Europe. Sugar technologists often use an old statement regarding a sugar factory’ s goal: “The sugar factory does not make sugar; it separates non sugars.” But this is only half a picture because today’s beet-sugar factory cannot survive if it doesn’t aim for the following:

- To produce sugar with the highest efficiency. - To produce molasses with the lowest possible purity. - To be responsive to the natural environment with its wastes."[1] 3

Table 1.1 Annual Consumption of Sugar in The World

Country India European Union China Brazil United States Other Total

[9]

2007/08 2008/09 2009/10 2010/11 2011/12 2012/13 22,021 16,496 14,250 11,400 9,590 77,098 150,855

23,500 16,760 14,500 11,650 9,473 76,604 152,487

22,500 17,400 14,300 11,800 9,861 77,915 153,776

23,500 17,800 14,000 12,000 10,086 78,717 156,103

25,500 17,800 14,400 11,500 10,251 80,751 160,202

26,500 17,800 14,900 11,700 10,364 81,750 163,014

"Today, the annual consumption is running at about 160 million tons and is expanding at a rate of about 2 million tons per year. The European Union, Brazil and India are the top three producers and together account for some 40% of the annual production. However most sugar is consumed within the country of production and only approximately 25% is traded internationally. Sugar cane is cultivated in over 100 countries and the amount of sugar from sugar cane is approximately 6 times higher than of beet sugar."[8]

Figure 1.1 Beet and Cane Distribution in The World [8]

"The word sugar comes from the Indian "sarkara". The chemical name of sugar is "sucrose" in English and "saccharose" in some European languages. Sugar (sucrose, C12H22O11) is one of the families of sugars (saccharides). All sugars belong to a larger group, known as carbohydrates (sugars, starches, and dietary fibers). The term sugar substitutes refers to all natural and 4

synthetic (artificial) sugars other than sucrose. All sweet-taste sugars and sugar substitutes are known to us as sweeteners. Sucrose (sugar), glucose (dextrose), and fructose (levulose) are examples of sweet-tasting sugars. The quantity of hydroxyl groups (OH) in molecules of sugars contributes to their sweetness. However, not all sugars are sweet in taste. In general, sugars with at least two hydroxyl groups (OH) in their molecules are sweet. About 50 compounds have a sweet taste. Sugar in its market-quality form is white and crystalline (granulated) with a sweet taste. It is used in the kitchen, as an ingredient in sugar-added food products (e.g., soft drinks and confectioneries), and in production of nonfood products (e.g., detergents and ethanol). Sugar is considered as: 

one of the world’s purest food products (99.95% sucrose; water is the main

remainder). 

One of the most natural foods used by humans for centuries without proven health risk.



One of the most purified organic compounds in the world and one of the lowest priced food products."[1]

1.2 History of Beet-Sugar "Sugarcane cultivation and the technique of sugar production began in India probably around 2000 BC and moved to Persia (now Iran) around AD 600. In Persia, the technique was improved; milk was used as the purifying agent; and the filtered syrup was crystallized. Then, the Persians invented a coneshaped clay mold for the production of cone sugar (loafsugar). In AD 800, sugarcane cultivation spread from Persia to Egypt, Syria, and as far as Morocco and Spain. By the fourteenth century, Egypt was Europe’s main supplier, via the port of Alexandria, of sugar made from sugarcane. Sugar became popular in tea in Britain by the end of the seventeenth century. In those days, sugar was available in large cone shapes that had to be broken first into large pieces with a cast-iron pincer and then into regular cube size with a little chopper. Sugarbeet cultivation on a research scale began in 1747 when Andreas Marggraf (a German chemist) discovered sugar in sugar-beet varieties. Later, 5

Franz Achard (Marggraf’s student) in Germany and Esipov in Russia were simultaneously engaged in the cultivation of sugar-beet varieties. They also continued independently with research on the processing of sugar from sugarbeet in industrial scale. The first beet-sugar factory was built in Cunern (in Germany) in 1802 by Achard and in Alyabevo (in Russia), shortly thereafter. Beetsugar technology developed rapidly, resulting in more than 400 beetsugar factories in European countries by 1830. During the industrial revolution between the late eighteenth and midnineteenth centuries, sugar technologists had always been the models for other industries by developing new technologies and equipment."[1]

1.3 By – Products of The Beet-Sugar Factory "In 100 kg sugar-beet, about 75 kg is water and 25 kg is dry substance. About 15 kg of the dry substance contained in 100 kg beet ends up in the form of a product (the amount of sugar produced is about one-seventh the weight of the t he beets), and the remaining 10 kg is a combination of by-products. The by-products of the beet-sugar factory are the following: - Pulp - Carbonation-lime residue (CLR) - Molasses 1.3.1 Pulp

Pulp which are usually dried and processed as animal feed. The dried pulp is often pelletized to increase bulk density to minimize shipping costs to distant markets. The dried pulp is valuable cattle feed because it supplies carbohydrates, proteins, and minerals. 1.3.2 Carbonation-Lime Residue

Sugar factories use lime, in the form of calcium hydroxide [Ca(OH)2, known as milk of lime] in the juice-purification station, to improve the quality of the beet juice. After the lime has been used, the juice is mixed with carbon dioxide gas (CO2), which precipitates lime again in the form of calcium carbonate (CaCO3). The calcium carbonate is then concentrated using cake filters

6

(rotary drum filters or filter presses) to produce carbonation-lime residue (CLR). The CLR stockpiled in the factory can take up space. Therefore, the sugar industry has been trying to find different uses for this material. Carbonationlime residue consists of about 80% CaCO3 as dry substance. It is ideal for use as an agricultural soil-enhancer and pH adjuster. 1.3.3 Molasses

Molasses is the runoff syrup from the final stage of crystallization, at which stage further separation of the sugar is not possible with conventional equipment. Beet molasses must not contain less than 48% sugar (namely sucrose and invert sugar) and 79.5% dry substances when is offered to consumers. Molasses produced by beet-sugar factories usually contains about 50% sugar and 80%dry substances. It is the most valuable by-product of the sugar factory. Its market includes the yeast fermentation, pharmaceutical, and animal feed industries."[1] 1.4 Sucrose Properties 1.4.1 Sucrose Formation

"Sucrose is formed in many fruits. Industrially, sugar beet and sugarcane are the main sources of sucrose. Sucrose is formed by the photosynthetic process .The primary product of photosynthesis is monosaccharides. Then, two monosaccharides combine with the help of an enzyme (named sucrase, not sucrose) to form sucrose: 6 CO2 + 6 H2O + Sunlight energy C6H12O6 + C6H12O6 + Enzyme

 

C6H12O6 + 6 O2 C12H22O11 + H2O

The second equation is called a dehydration reaction (condensation reaction) because one molecule of water is formed f ormed during reaction. Sucrose is found in many fruits (about 7% in bananas, 5% in oranges, and to a lesser extent in some vegetables). Sugar beet and sugarcane are the main sources of sucrose. Sugar beet contains 15 to 20% and sugarcane 10 to 17% sucrose."[1]

7

1.4.2 Sucrose Properties Table 1.2 Sucrose Properties



Chemical Formula

C12H22O11

Density (Kg/m3)

1587.1

Melting point (oC)

185

SUCROSE SOLUBILITY

"Sucrose is highly soluble in water. The reason for its high solubility is that sucrose contains eight free hydroxyl groups (OH) of which at least five can attach to other molecules depending on the molecule reactivity. When sucrose is dissolved in water, each sucrose molecule can form at least five hydrogen bonds with water molecules. As the concentration of the solution increases, no more free water molecules are left in the solution for hydrogen bonding and the solution becomes saturated. Solubility increases with increasing temperature."[1]

Figure 1.2 The Relationship Between Solubility with Temperature[1]

"Sucrose (a non polar substance) is soluble in polar solvents, such as water, acetone (C3H6O), acetic acid (HCH3COO), and formic acid (HCOOH). But it is not soluble in nonpolar solvents, such as chloroform (CHCl3), carbon tetrachloride (CCl4), hexane (C6H14), heptane (C7H16), toluene (C7H8), oil."[1]

8



SUCROSE SPECIFIC HEAT

"The specific heat of pure and impure sucrose solutions decreases when their concentration and temperature are increased. You can calculate the specific heat of pure and impure sucrose solutions by using equation (1.1) :

CP = 4.187 − DS × (0.0297 − 4.6 × 10 − × P ) + 7.5 × 10− × DS × T 5

5

(1.1)

DS: Dry substance content (for pure sucrose solutions, DS = S) T: Temperature (ºC) P: Purity (for pure solutions, P = 100)"[1]



SUCROSE SPECIFIC ENTHALPY:

"The specific enthalpy (in kJ/kg) of pure and impure sucrose solutions using equation (1.2) :

H=

T {4.187 − DS ( 0.0297− 4.6 × 10− × P ) +3.75 × 10− × DS × T} 5

5

(1.2)

DS: Dry substance content (for pure sucrose solutions, DS = S) T: Temperature (ºC) P: Purity (for pure solutions, P = 100)"[1]

1.4.3 Sucrochemistry:

"Sucrochemistry is the study of using sucrose as a raw material for production of other products. Many sucrose-based products are available on the market, such as the following: 

BIOETHANOL

Ethanol produced from plant sources produced from sucrose, molasses, or sucrose containing raw materials (mainly sugarcane) is the most important nonfood product of sucrose. Ethanol-based fuel is considered as the gasoline 9

of the future. Because of a recent steady increase of oil price, every country tries to reduce its dependence on oil-based fuel. In the United States, a program called 25/25 aims to get 25% of its energy from renewable resources (e.g., ethanol, solar, and wind) by the year 2025. Ethanol produced from sugarcane and corn plays an important role in this program. Bioethanol is produced in countries such as Brazil, the United States (ethanol is produced from corn), India, Pakistan, and some European countries. It is mixed up to 85% with gasoline as a blend of fuel for most cars, or it is used without mixing in specially designed cars. Bioethanol consumption counts for 2.5% of the world gasoline use In the world, Brazil leads sugar and ethanol production, both of which come from sugarcane. In Brazil, ethanol is produced from sugarcane juice, molasses, or mix of both with an expected production of about 17.5 million m3 in 2005/2006 campaign. As a result, Brazil became an energy-independent nation and its sugar surplus was reduced. Ethanol produced in Brazil costs less than U.S. ethanol produced from corn. European countries have become interested in ethanol production from sugar beet. France produces 125000 m3 ethanol a year from sugar beet, wheat, and maize.



DEXTRAN

A polysaccharide used in pharmaceutical and photographic industry, and some other polysaccharides. Dextran is produced through a fermentation process. Sucrose is also used for the production of products such as acetic acid (HCH3COO).

10



CARBOHYDRATES:

Carbohydrates (from carbon hydrates) are saccharides, starches, and dietary fibers. Carbohydrates are a large class of substances with similar structures.



MONOSACCHARIDES:

Monosaccharides are the simplest carbohydrates consisting of one unit, so they cannot be further decomposed to a simpler sugar. Example: Glucose (blood sugar), Fructose (fruit sugar).



DISACCHARIDES:

Disaccharides are carbohydrates composed of two monosaccharides. Example: Sucrose (sugar), lactose (milk sugar), and maltose (malt sugar).



OLIGOSACCHARIDES:

Oligosaccharides (from the Latin oligo meaning a few) consist of three to six monosaccharides. Example: Stachyose, Raffinose .



POLYSACCHARIDES:

Polysaccharides (complex carbohydrates) are long molecules that consist of many repeating monosaccharides (hundreds to thousands) connected by alpha 1 : 2 glycosidic bonds. Example: Starch, Pectin, Cellulose, Dextran."[1]

11

Chapter 2. Process Description

"The agricultural department of the sugar factory is responsible for coordinating with sugarbeet farmers the delivery, unloading, and storing of sugarbeets, as well as care of sugarbeets during storage. At the point where the factory starts to process sugarbeets, responsibility passes to the operation department, which must recover as much sugar as possible from the sugarbeets entered into the process. Changing sugarbeet to sugar requires successive processes, called unit operations. Unit operation is one of the subjects that differentiates chemical engineering from chemistry at universities. Unit operation studies focus on the chemical processes used in chemical and food factories. Sugar production is a large-scale operation divided into smaller unit operations. To process sugarbeet and produce sugar, unit operations are commonly used in a beet sugar factory: Slicing, Diffusion, Purification, Sedimentation, Filtration, Evaporation, Crystallization, Centrifugation, Drying."[1]

2.1 Block Diagram :

Figure 2.1 Block Diagram of Process

[1]

12

2.2

Flow sheet of Process :

The process retainer for the project is shown in Figure. (2.2)

14

4 Diffusion Tower

7 2 1

3 Mixer

5

6

L i m i n g

8

14

13

11

Evaporator

10

Filter

Clarifier

12

9

15

S u l f i t a t i o n

17 16

18

19

Centrifuge

Crystallizer

20

21 25

24

23

Storage

Dryer

22

26

Figure 2.2 Flow sheet of Process

13

In flow sheet, we have 15 main units of the process, (Mixer, Diffusion Tower, Heater1, Reactor, Clarifier, Heater2, Filter, Evaporator, Tank, Crystallizer, Cooler1, Centrifuge, Dryer, Cooler2 and Storage).

2.3

Description of Flow sheet

Figure (2.2) shows the sliced beet (cossettes) in stream 1 enters a mixer, and the diffusion juice in stream 3 will be mixed with cossettes , after mixing operation the thin juice (cossettes + diffusion juice) in stream 2 enters the diffusion tower and the mass transfer operation will be between (thin juice in stream 2 and the condensate water coming from evaporator in stream 14) , the pulp will exit as stream 4 , and the juice returns to the mixer and exits , then it is heated from 30 OC to 88 OC before entering the liming unit (stream 6) . In liming unit there are two steps : The first is the addition Milk of Lime (MOL) Ca(OH)2 to remove the impurities in juice after mixing operation. The second is Carbonation Process (Adding CO2) to precipitate CaO from juice. Stream 8 (thin juice + precipitated CaO) enter clarifier to remove CaO precipitated in Carbonation Process. Then the clarifying juice (Stream 10) is heated to 92OC before entering the filter to remove non sugars (increasing purity of sucrose in dry substance from 88% to 92%). The thin juice in previous units has (15% dry substance and 85% water). In evaporation unit the concentration of dry substance will be increased to 70% , water 30% in stream 15 with temperature about 95OC before adding SO2 to decolorize it from impurities with thick juice. Then, the thick juice in stream 16 enters the crystallizer to form the crystals of sugar and the dry substance here will be 93% and 7% water. Then the wet crystals will be cooled to 75OC before entering the centrifuge unit (stream 19) which separate the crystals from molasses by centrifuging force. The molasses here (stream 20) is available as a product for selling and the wet crystals (stream 21) will enter the dryer and leave it with 100% sucrose and 0% water before cooling to 30OC (stream 24). The product (white sugar) now is available in storage (silos) at 30OC with 100% sugar but different shape (granulated sugar in stream 25 and powder of sugar in stream 26).

14

Chapter 3. Material Balance 3.1

The General Equation of Mass Balance :

The aim of this chapter is to carry out the mass balance calculation about the giving flow sheet in Fig.(2.2). The general equation of mass balance calculation will be considered as follows: [2]

  =    

(3.1)

Accumulation in process equal zero . Mass Input = mass flow rate streams enter system. Mass Output = mass flow rate streams leave system. In this project, we assume 300 ton/day as our product, and the factory works 180 days. TABLE. (3.1) Materials used in the process and their abbreviation

Name

Species

Symbol

Water

H2O

W

Sugar

C12H22O12

S

Non Sugar

Mg,Ca,….

NS

- The amount of Beet entering to the process after washing and slicing (Cossetts) is 3554 ton/day (by backward mass balance). - Factory works 180 days. - m1= 148083 kg/h (the amount of beet) - m4= 29616 kg/h (the amount of pulp) - m7= 5574 kg/h (the amount of lime added to the reactor) - m20 = 19737 kg/h (the amount of molasses produced)

15

3.2

Mass Balance of Units :

Water

14 1

Overall mass Beet

5

balance of Diffusion

(C12H22O12) Juice diff.

XDS = 0.15 Xw = 0.85

4

Pulp Xp= 0.2 on beet

  =   Overall M.B. : m1+ m14= m4 + m5

 m1+

3845.87= m4 + 6689.05

m4=0.2m1  m1+3845.87=

0.2m1 + 6689.05

 0.8m1=6689.05-3845.87  m1=3554  m4=

ton/day = 148083.33 kg/h

(0.2)(3554)=710.8 ton/day = 29616.67 kg/h

Component M.B. : STREAM 5 : m5DS = 0.15 (6689.05) = 1003.36 ton/day =41806.67 kg/h m5S = 1003.36 (0.88) = 883 ton/day = 36791.67 kg/h m5NS = 1003.36 (0.12) = 120.36 ton/day = 5015 kg/h m5w = 0.85 (6689.05) = 5685.7 ton/day = 236904.2 kg/h 16

Mixer – Unit 1 : The cossette mixed with diffusion juice .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

2

226567.5

35184.6

4797.9

3

333005.34

52427.5

7149.17

5

236904.2

36791.67

5015

Diffuser – Unit 2 : To extract sucrose from cossette by using water .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

2

226567.5

35184.6

4797.9

3

333005.34

52427.5

7149.17

14

160244.8

0

0

o

Heater – Unit 3: In this unit the juice will be heated from 30 to 88 C .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

5

236904.2

36791.7

5015

6

236904.2

36791.7

5015

Liming – Unit 4 : Liming is the step of purification where lime is added to the

juice .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

6

236904.2

36791.7

5015

8

241641.9

37525.4

5117.1

17

Clarifier – Unit 5 : In this unit the lime will be separated from juice .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

8

241641.8958

37525.4

5117.1

9

21321.375

9380.6167

1279.175

10

181233.95

28144.7833

3837.925 o

Heater – Unit 6 : In this unit the juice will be heated from 88 to 90 C .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

10

181233.95

28144.7833

3837.925

11

181233.95

28144.7833

3837.925

Filter – Unit 7 : In this unit the purity increased to 90 .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

11

181233.95

28144.8

3837.925

12

7878.4

0

1390.3125

13

173355.7

28144.8

2447.375

Evaporator – Unit 8 : In this unit the dry substance increased from 15 to 70.

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

13

173355.7

28144.8

2447.375

14

160244.8

0

0

15

13110.9

28144.8

2447.375

18

Sulfitation – Unit 9 : In this unit added SO2 to remove color .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

15

13110.9

28144.8

2447.375

16

13110.9

28144.8

2447.375

Crystallizer – Unit 10 : In this unit the crystals of sugar will be formed .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

16

13110.9

28144.8

2447.375

17

10809.4

0

0

18

2301.6

28144.8

2447.375

o

Cooler – Unit 11: In this unit the juice will be cooled from 100 to 75 C .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

18

2301.5625

28144.8125

2447.375

19

2301.5625

28144.8125

2447.375

Centrifuge – Unit 12: In this unit,the sugar crystals are separated from molasses

.

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

19

2302.56

28144.33

2447.33

20

1184.25

9831.74

854.933

21

657.5

12500

0

19

Dryer – Unit 13 : In this unit , the wet crystal entered the dryer and leave it as

dry crystal ( W= 0% , S = 100% )

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

21

657.5

12500

0

22

657.5

0

0

23

0

12500

0

o

Cooler – Unit 14 : In this unit the sugar will be cooled from 65 to 30 C

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

23

0

12500

0

24

0

12500

0

o

Storage – Unit 15 : To keep sugar at 25 C .

Stream

Water(kg/h)

Sugar(kg/h)

Non Sugar(kg/h)

24

0

12500

0

25

0

6250

0

26

0

6250

0

20

Chapter 4. Energy Balance The aim of this chapter is to carry out the energy balance calculation around units in the flow sheet given in Fig.(2.2). The general equation of heat balance calculation will be considered as follows:[2]

 =   

(4.1)

Accumulation in process equal zero

 =    −   

(4.2)

We assume 25 oC as a reference temperature for our calculations.

4-1

Heat Capacities :

CpiW = 75.4 * 10-3 kJ/kmol.oC and CpiLime= 89.5*10-3 kJ/kg.oC[2] CpiDS From Appendix D.1 attached.

21

Mixer – Unit 1

2

Beet T= 25oC H1

(C12H22O12) T= 65oC H3

Mixer

1

3

(C12H22O12) T= 70oC H4

5

(C12H22O12) T= 35oC H5

Water @ 25oC

H1=m1cp ∆T



Cp = Ai + BTi + CT2 ( kJ/kg .oC) = 75.4*10-3 kJ/kmol.oC

∆T = 25-25=0  H1=0

H2 = H2DS + H2W H2DS = m2DS CpDS ∆T CPDS at 65oC (90% purity)(15%DS) = 3.8775 kJ/kg.oC (From appendix D.1). H2DS = (959.58 *103)( 3.8775)(65-25) = 148830858kJ/day = 6201285.75 kJ/h   .  H2W =   CPdT = [75.4*10-3] [65-25]







= 9111034440 kJ/day = 379626435 kJ/h H2 = 9259865298 kJ/day = 385827720.8 kJ/h H3 = H3DS + H3W H3DS = m3DS CpDS ∆T CPDS at 70oC (90% purity)(15%DS) = 3.88 kJ/kg.oC (From appendix D.1). H3DS = (1429.84*103)(3.88)(70-25) = 249650064 kJ/day = 10402086 kJ/h

22

  .  -3 H3W =  C dT = [75.4*10 ] [70-25] P  



= 1527308055 kJ/day = 63637835.63 kJ/h H3 = 1776958119 kJ/day = 74039921.63 kJ/h H5 = H5DS + H5W H5DS = m5DS CpDS ∆T CPDS at 30oC (90% purity)(15%DS) = 3.835 kJ/kg.oC (From appendices attached) H5DS = (1003.36*103)(3.835)(30-25) = 19239428 kJ/day = 801642.83 kJ/h   .  -3 H5W =  C dT =  P  [75.4*10 ] [30-25]



= 137933618.3 kJ/day = 5747234.1 kJ/h H5 = 157173046.3 kJ/day = 6548876.93 kJ/h Q = ∑Hout - ∑Hin = (H5 + H2) – (H1+H3)

Q = 7640080225 kJ/day = 318336676.1 kJ/h

23

Heater – Unit 3 Temperature(T)

Heat capacity of Dry (CP)

Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

5

30

3.835

6548876.93

6

88

3.905

72801444.38

Streams

Amount of heat (Q) = 66252567.46KJ/h

Diffuser – Unit 2

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

2

65

3.8775

385827720.8

3

70

3.88

74039921.63

4

70

3.88

2174126.11

14

90

-

43622136.58

Streams

Amount of heat (Q) = -353235809.6 KJ/h

Liming – Unit 4

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

6

88

3.905

72801444.38

7

40

-

100989.46

8

88

3.905

74260150.38

Streams

Amount of heat (Q) = 1357716.54 KJ/h

24

Clarifier – Unit 5 Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

8

88

3.905

74260150.38

9

88

2.87

14621985.76

10

88

3.905

55695000.60

Streams


Heater – Unit 6

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

10

88

3.905

55695000.60

11

92

3.908

59237619.49

Streams


25

Filter – Unit 7

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

11

92

3.908

59237619.49

12

92

3.908

2575166.572

13

92

3.908

56663384.47

Streams


Evaporator – Unit 8

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

13

92

3.908

56663384.47

14

100

----

50343572.05

15

95

2.985

10236656.06

Streams


26

Crystallizer – Unit 10 Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

16

95

2.985

10236656.06

17

100

------

81494833.33

18

100

2.43

12809018.75

Streams


Cooler – Unit 11

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

18

100

2.43

12809018.75

19

75

2.39

8478162.50

Streams


27

Centrifuge – Unit 12

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

19

75

2.39

555362

20

60

2.175

1412539.78

21

60

-----

949354.23

Streams


Dryer – Unit 13

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

21

60

----

949354.23

22

65

----

110167.78

23

65

----

927000

Streams

Amount of heat (Q) = 87813.55 KJ/h Cooling tower – Unit 14

Temperature(T)


Total Enthalpy(HT)

(OC)

(KJ/KgoC)

(KJ/h)

23

65

----

927000

24

30

----

106195.31

Streams

Amount of heat (Q) = -1082250 KJ/h

28

Chapter 5. Equipment Design 5.1

Reactor Design (Unit – 4) :

Carbonation is the step of purification where the carbonation gas is added to the limed juice. The gassing time of the juice in this step is about 10 minutes.[1] The carbonation has main function is to react with unreacted lime to precipitate calcium carbonate.

C12H22O11.CaO + CO2

C12H22O11 + CaCO3

(5.1)

Assumptions : - Operating at steady state and the reactor is isothermal . - Negligible heat resistance between the pellets and gas. - The pressure drop along the reactor is negligible.

B= C12H22O11.CaO

C12H22O11

XB=0.98 T=88oC P=2 atm

Reactor

A=CO2 CaCO3

X A=0.02 Figure 5.1 Reactor Design

29

- A is limiting of reaction .

To calculate the volume of reactor, the equation (5.2) shown is :[3]

(5.2) The percentage of CO2 in the carbonation gas entering the carbonation tank (CO2)IN to CO2 in the gas leaving the tank (CO 2)OUT , It can be calculated by the following formula :[1]

X=

( (  (    –  10000

(5.3)

CO2(IN) = 2  of lime juice . CO2(out)= 0.4  of lime juice .

X=

. . 10000=0.8

m8 =

6822.83 ton/day = 284284.6 kg/h

 m(CO2)in=

(284284.6)(0.02)=5685.69 kg/h

The rate of reaction is given by : [3]

− = 

(5.4)

Where:

 =  ε= y AoԺ

0

 

(5.5)

 Ժ=1+1-1-1=0

(5.6) 30



ε= 0

 =  0  0

 = 0

(5.7)

(. .( -3 3 3 = 1.3510 Kmol/m = 1.35 mol/m (.  . (

C A = 1.35(1-0.8) = 0.27 mol/m3 



=  =  = 0.98/0.02= 49

 = (  − ( (  

(5.8)

(5.9)

CB= C Ao (49 – 0.8) = 1.35(48.2) = 65.07 mol/m3

Kinetic :[3]

 =   

(5.10)

E = activation energy = 50000 J/mol R = gas constant = 8.314 J/mol.k A = frequency factor = 3.29105 T= absolute temperature = 361 K k=3.29X105 e(-50000/(8.314)(361)) k=0.0192 S-1 Now, Applying equation (5.4) : -r A= KC ACB -r A = (0.0192)(65.07)(0.27) = 0.34 mol/m3.s

31

(. . (  FBo=   ( ( FBo= 129.22 kmol/h = (129.22)(1000/3600) = 35.89 mol/s V= (35.8945)(0.8)/(0.34) = 84.46 m3 The volume is big , then will dived to two reactor (42.23 m3 per reactor).

Table 5.1 Reactor Design

5.2

Temperature (T)

88oC

Pressure (P)

2atm

Concentration of A (C A)

0.27 mol/m3

Concentration of B (CB)

65.07 mol/m3

activation energy , E

50000 J/mol

frequency factor, A

3.29105

k

0.0192 S-1

Number of Reactors

2

Volume of one reactor needed

42.23 m3

Filter Design (Unit – 7) :

Filtration is the process of separating solid particles from a liquid by the force of the pressure difference on both sides of the filter medium (a porous material). The filter residue is the solid that remains on the filter medium (nonsugars), and filtrate (thin juice) is the liquid that passes through the filter medium and is collected during filtration. There are many types of filters , we will use Rotary Drum Filter in our project because the amount of cake on filter is very big with time , and the rotary drum can stand this large amounts. Type : Rotary drum filter (continuous filtration).

32

Figure 5.2 Rotary Drum Filter

The general relation to specify the total area of rotary drum filter "with constant pressure drop" :[5]

        = (   

(5.11)

Assumptions : - The pressure drop will be constant in rotary drum filter , ∆P = 103.42 KPa - The cake resistance (α 0 = 2.7 x 1010 m/kg) - The compressibility coefficient of the cake (s = 0.29) - The cycle time (Tc = 4 minutes = 4 x 60 = 240 second) - The submergence fraction (f = A/AT = 0.4)

The total area of filter can be calculated using equation (5.12) :[5]

 =

     (     

(5.12)

MF from mass balance = 213218.1 Kg/h = 59.22 Kg/s Mc from mass balance = 3837.925 Kg/h = 1.066 Kg/s MF/MC = 55.5

33

ρsugar = 1587 Kg/m3

C = 15 / [1 – (mF/mc – 1) 15/ρsugar ] = 33.35 Kg/m3 µw at (92oC) = 0.3095*10-3 Kg/m.s

Hence, the total area : AT = 1.066 [(2.70*1010 )(0.3095*10-3)]/[2*33.35*10342140.71*0.4*(1/240)]0.5 AT = 67.7 m2 π*D*L = 67.7 D*L= 67.7/ π = 21.5

In filter design , the length at least double of diameter , hence, the diameter D about 3 m and length L about 7.2 m . Submerged area (A) = f * AT  0.4

* 67.7 = 27.08 m2 . Table 5.2 Filter Design

Pressure drop (∆P)

103.4214 KPa

Cake resistance (α 0)

2.7*1010 m/Kg

Compressibility coefficient (s)

0.29

The cycle time (Tc)

4 minutes (240 second)

The submergence fraction (f)

40 % (0.4)

Total area of filter (AT)

67.7 m2

The diameter (D)

3m

The length (L)

7.2 m

The submergence area (A)

27.08 m2

34

5.3

Evaporator Design (Unit – 8) :

Evaporation is the process of concentrating a solution by boiling to convert some of the liquid to vapor. The evaporation station can be viewed as the heat center of the factory because it serves the sugar factory in three ways: - Concentrates thin juice and produces thick juice. - Produces condensate for boilers. - Produces vapor and condensate for heat users. In our project, multiple effect evaporator is used.[1]

Figure 5.3 Multiple Effect Evaporator [10]

Type : (Multiple Effect Evaporator).

Assumptions : - Number of effects = 3 - U1 = 3123 , U2 = 1987 , U3 = 1136 W/m2.oC [4] - Boiling Point Rise (BPR OC) = 1.78DS + 6.22DS2 (DS = % sugar in each effect) [4] - Cp = 4.19 – 2.35DS KJ/kg.OC [4] - Each effect has same surface area - Each effect has same amount of water vaporized (V1 = V2 = V3) - The vapor pressure at third effect is 84.55 KPa - The temperature of saturation steam is 180OC

35

STEP 1 : The temperature of the third effect at 84.55 KPa is 95 OC , the temperature of the thick juice can be estimated as follows : BPR3 = 1.78 (0.7) + 6.22 (0.72) = 4.2938OC T3 = 95 + 4.2938 = 99.2938 OC STEP 2 : From overall mass balance in each effect : 1st effect with (V1 = V2 = V3 = 160244.79 / 3) : F = V1 + L1  L1 = 203947.9167 – 53414 .93 = 150532.9861 Kg/h 2nd effect : L1 = V2 + L2  L2 = 150532.9861 – 53414.93 = 97118.05613 Kg/h 3rd effect : L2 = V3 + L3  L3 = 97118.05613 – 53414.93 = 43703.126 Kg/h From component mass balance in each effect : 1st effect with (V1 = V2 = V3 = 160244.79 / 3) : 0.15 F = L1x1  30592.1875 = 150532.9861 x1  x1 = 0.203 2nd effect : 0.203L1 = L2x2  30558.196 = 97118.05613 x2  x2 = 0.314 3rd effect : 0.314L2 = L3x3  30495.06962 = 43703.126 x3  x3 = 0.70 STEP 3 : The BPR in each effect : BPR1 = 1.78 (0.203) + 6.22 (0.2032) = 0.617OC BPR2 = 1.78 (0.314) + 6.22 (0.3142) = 1.172OC BPR3 = 1.78 (0.7) + 6.22 (0.72) = 4.2938OC ∑∆T = TSTEAM – T3Sat - ∑BPR

(5.13)[4] 36

= 180 – 95 – (0.617 + 1.172 + 4.2938) = 78.9172OC ∆T1 = ∑∆T [(1/U1)/(1/U1+1/U2+1/U3)]

(5.14)[4]

= (78.9172) [(1/3123)/(1/3123+1/1987+1/1136)] = 14.831OC ∆T2 = ∑∆T [(1/U2)/(1/U1+1/U2+1/U3)]

= (78.9172) [(1/1987)/(1/3123+1/1987+1/1136)] = 23.311OC ∆T3 = ∑∆T [(1/U3)/(1/U1+1/U2+1/U3)]

= (78.9172) [(1/1136)/(1/3123+1/1987+1/1136)] = 40.774OC Actual BP in each effect :[4] T1 = TS1 - ∆T1 = 180 – 14.831 = 165.169OC = TS2 T2 = T1 - ∆T2 – BPR1 = 165.169 – 23.311 – 0.617 = 141.241OC = TS3 T3 = T2 - ∆T3 – BPR2 = 141.241 – 40.774 – 1.172 = 99.3OC STEP 4 : The heat capacity of the juice in each effect (4.19 – 2.35DS) : F : Cp1 = 4.19 – 2.35 (0.15) = 3.83 KJ/Kg.OC L1 : Cp2 = 4.19 – 2.35 (0.203) = 3.71 KJ/Kg.OC L2 : Cp3 = 4.19 – 2.35 (0.314) = 3.45 KJ/Kg.OC L3 : Cp4 = 4.19 – 2.35 (0.70) = 2.54 KJ/Kg.OC Enthalpy in each effect : [4] Effect (1) : H1 = HS2 + 1.884 (BPR1) = 2763.5 + 1.884 (0.617) = 2764.6624 KJ/Kg λs1 = HS1 – hS1 = 2778.2 – 763.22 = 2014.98 KJ/Kg

Effect (2) : H2 = HS3 + 1.884 (BPR2) = 2735 + 1.884 (1.172) = 2737.208 KJ/Kg λs2 = HS2 – hS2 = 2763.5 – 697.34 = 2066.16 KJ/Kg

Effect (3) : H3 = HS4 + 1.884 (BPR3) = 2668.1 + 1.884 (4.2938) = 2676.186 KJ/Kg λs3 = HS3 – hS3 = 2735 – 592 = 2143 KJ/Kg

Heat balance in each effect , noting this relations : 37

V1 = 203947.9167 – L1 V2 = L1 – L2 V3 = L2 – 43703.126 Effect (1) : FCp1(TF – 0) + S λs1 = L1Cp2(T1 – 0)+ V1H1 

203947.9167 (3.83) (92) + S (2014.98) = L1 (3.71) (165.169) +

(203947.9167 – L1) (2764.6624) Effect (2) : L1Cp2(T1 – 0) + V1 λs2 = L2Cp3(T2 – 0)+ V2H2  L1 (3.71)

(165.169) + (203947.9167 – L1) (2066.16) = L2 (3.45) (141.241) +

(L1 – L2) (2737.208) Effect (3) : L2Cp3(T2 – 0) + V2 λs3 = L3Cp4(T3 – 0)+ V3H3 

L2 (3.45) (141.241) + (L1 – L2) (2143) = 43703.126 (2.54) (99.3) +

(57461.5182) (2676.186)

Solving equation for effects (2) and (3) simultaneously : L1 = 155188.6455 Kg/h ,

V1 = 48759.2712 Kg/h

L2 = 101164.6418 Kg/h ,

V2 = 54024.0037 Kg/h

L3 = 43703.126 Kg/h

, V3 = 57461.5182 Kg/h

From effect (1) : (203947.9167) (3.83) (92) + S(2014.98)=(155188.6455)(3.71)(165.169)+ (48759.2712) (2764.6624) The amount of steam used in first effect : S = 78430.489 Kg/h

38

STEP 5 : The heat required of each effect and area can be calculated as follows : [4]

 =  

(5.15)

Q1 = SλS1 = (78430.489/3600) * (2014.98*1000) = 43.898*106 W Q2 = V1λS2 (48759.2712/3600) * (2066.16*1000) = 27.984*106 W Q3 = V2λS3 (54024.0037/3600) * (2143*1000) = 32.159*106 W Hence, the area required for each effect from equation (5.16) shown: [4]

 =

  

(5.16)

A1 = (Q1)/(U1*∆T1) = (43.898*106)/(3123*14.831) = 947.76 m2 A2 = (Q2)/(U2*∆T2) = (29.661*106)/(1987*8.541) = 640.36 m2 A3 = (Q3)/(U3*∆T3) = (33.239*106)/(1136*14.94) = 717.60 m2

Table 5.3 Evaporator Design

The amount of steam at 1st effect (S)

78430.489 Kg/h

The amount of heat at 1st effect (Q1)

43.898*106 W

The amount of heat at 2nd effect (Q2)

27.984*106 W

The amount of heat at 3rd effect (Q3)

32.159*106 W

The total area in 1st effect

947.76 m2

The total area 2nd effect

640.36 m2

The total area in 3rd effect

717.60 m2

The temperature of thick juice (TOUT)

95 OC

The temperature of steam (TS)

180 OC

39

5.4

Dryer Design (Unit – 13) :

Drying is the process of removing a liquid (usually water) from a wet solid substance by the use of heat (usually hot air). The concentration difference between the heating medium and the solid substance under drying is the driving force (cause) of drying, which evaporates the water in the substance. In normal operations, if the sugar entering the dryer has a moisture content below 1% and is at about 60 C, its heat provides sufficient energy to reduce the moisture to about 0.00%. Most of the water in wet sugar is on the surface of the crystals in the form of a saturated sucrose solution with high purity (very close to 100%) and high concentration (about 90%). The wet sugar is dried and cooled in the sugar dryer to a moisture content of about 0.00% and a temperature below 30˚C. The sugar loses more moisture and cools during

transportation from the dryer to the silo or in the packing station. The dryer used is rotary drum because the efficiency is higher than other types.

Type : Rotary drum dryer.

Figure 5.4 Rotary Drum Dryer [1]

Assumption : Mass velocity of dry air (G) = 2000 Kg/m2.h Temperature of entering air (Thb) = 126.7°C Temperature of exiting air (Tha) = 70°C The amount of mass entering the dryer (m) = 13157.5 Kg/h Initial moisture (Xa) = 0.05 , Final moisture (Xb) = 0 , G = 2000 Kg/m2.h Tsa = 60°C , Tsb = 65°C , Thb = 126.7°C , Tha = 70°C , Csb = 0.57 kJ/Kg.°C Rate of mass transfer : mv = m(Xa – Xb) = 657.875 Kg/h , hb = 0.01

40

The heat duty was found from energy balance : Q = 87813.55 kJ/h The flow rate of entering air is found from heat balance :[5]

(1    = (  (5.17) (1    = 2717.087   = .  The outlet humidity : [5]

 =    =0.01 . . =.

(5.18)

The cross sectional area : [5]   .   = ( = =.  

(5.19)

The dryer diameter : [5]

.  = (   =.

(5.20)

The dryer length : [5] LMTD = Log Mean Temperature Difference

 = ..

(5.21)

. = . .. =.

41

Table 5.4 Dryer Design

mass entering the dryer (m)

13157.5 Kg/h

Initial moisture (Xa)

0.05

Final moisture (Xb)

0

Mass velocity of dry air (G )

2000 Kg/m2.h

Amount of heat (Q)

87813.55 kJ/h

The cross sectional area

. . .

The dryer diameter (D) The dryer length (L)

5.5

Cooling Tower Design (Unit – 14):

Cooling towers are heat removal devices used to transfer process waste heat to the atmosphere. Cooling towers make use of evaporation whereby some of the water is evaporated into a moving air stream and subsequently discharged into the atmosphere. As a result, the remainder of the water is cooled down significantly. The chosen cooling tower type that is natural draft cooling tower.

Figure 5.5 Natural Draft Cooling Tower [13]

42

Assumption : Temperature of entering water (t1) = 40 oC Temperature of exiting water (t2) = 25 oC Temperature of entering air (t1) = 20 oC Temperature of exiting air (t2) = 38 oC Overall heat transfer coefficient (U) = 650 w/m2.K The heat duty was found from energy balance : Q = - 820804.69 kJ/h

Mass flow rate of water : Cp,water at Tav= (40+25)/2= 32.5oC ,Cp,water = 4.18 (kJ/kg.oC) Δtwater = ( t2- t1) = (40 - 25) = 15oC

Q = mwater * Cpwater * ΔTwater = mwater * (4.18)(15) mwater =

13091 (kg/h)

Mass flow rate of air : Cp,air at Tav = (38+20)/2= 29 oC, Cp,air = 1.005 kJ/Kg.oC Δtair = ( t2- t1) = (38 - 20) = 18oC

Q = mair * Cpair * ΔTair = mwater * (1.005)(18) mair =

45373.39 (kg/h)

Design equation:

 =     

[5]

(5.22)

Where : U: overall heat transfer coefficient (w/m2.K). A: surface area for cooler (m2). ΔTln: log mean temperature difference.

43

ΔTln =

T ou t  T in

ln

T ou t

, ΔTout= (40-30)= 10oC, ΔTin= (65-25)= 40oC

T in

ΔTln=(10 - 40)/ln(10/40)= 21.64oC

Area of cooling tower = (Q/U* ΔTln)= (228001.303/(650*21.64))= 16.21 m2

Table 5.5 Cooling Tower Design

Mass flowrate of water (mwater )

13091 Kg/h

Mass flowrate of air (mair )

45373.39 Kg/h

Amount of heat (Q)

- 820804.69 kJ/h

Overall heat transfer coefficient (U)

650 w/m2.K

Log mean temperature difference (ΔT ln)

21.64oC

Area of cooling tower

16.21 m2

44

Chapter 6. Sizing equipment 6.1 Sizing of Mixer (Unit -1): Type : BMA (Braunschweig Machinenbau Anstalt). The cossette mixer works as a heat exchanger because the hot tower juice preheats the cossettes. The retention time = 15 minutes

  = . = 5.73 m /min   =  . 3

(6.1)[4]

V = 5.73* ( 15 ) = 85.95 m3 The volume is big , then will dived to three tanks (28.65 m3 per tank) Table 6.1 Mixer Sizing Diameter of one tank (D)

2.5 m

High of one tank (H)

5. 85 m

Retention time (TR)

15 minutes

Volumetric flow rate

5.73 m3/min

Number of tanks

3

Volume of one tank needed

28.65 m3

45

6.2 Sizing of Diffuser (Unit -2): Type : Braunschweig Machinenbau Anstalt (BMA) tower diffuser Assumptions : - The process takes about 50 minutes at 70ºC .

- A BMA diffuser (5-7) m in diameter and (11-16) m high can handle 5000 t beets/day .

  .   =  = . = 6.033 m /min 3

V = 6.033* ( 50 ) = 301.65 m3 Table 6.2 Diffusion Tower Sizing

Diameter (D)

5.44 m

High (H)

13 m

Time of process (t)

50 minutes


6.033 m3/min

Volume of diffuser needed

310.65 m3

46

6.3 Sizing of Heater (Unit-3): The area of heater is depending on overall heat transfer coefficient "U" (W/m2.oC). The value of "U" for organic compounds (sugar) with water is between (500 – 750). "U" = 700 W/m2.oC

The amount of heat required in heater is calculated from the general equation of heat exchanger (6.2): [4]

 =    ΔTlm =

T out  T in

ln

T ou t

(6.2)

(6.3)

T in

ΔTout = (170-88)= 82oC , ΔTin=(250-30)= 220oC ΔTlm = (82 – 220) / ln (82/220) = 139.83 oC

Qheater = 66252567.46 kJ/h (Appendix B.2) . Hence, the area of heater can be calculated now : A = (66252567.46 *103/3600) / (700*139.83) = 188 m2 Table 6.3 (Heater Sizing)


700 W/m2.OC

Logarithmic mean temperature ( ΔTlm)

139.83 OC

Amount of heat needed in heater (Q)

66252567.46 KJ/h

The area of heater (A)

188 m2

47

6.4 Sizing of Lime Unit (Unit - 4):

Liming is the step of purification where lime is added to the juice .

Type : BMA Vertical , H= (8-12) m , D= (2.5 – 4 ) m The retention time = 15 minutes

  = . = 4.26 m /min   =   3

V = 4.26 * ( 15 ) = 64 m3

Table 6.4 Lime Unit Sizing

Diameter (D)

3m

High (H)

9m

Retention time (TR)

15 minutes


4.26 m3/min

Volume of liming needed

64 m3

6.5 Sizing of Clarifier (Unit - 5): Type : Dorr clarifier . Vertical , H= (8 -12) m , D= 2-6 m The retention time = 20 minutes

  .   =  =  = 4.34 m /min 3

V = 4.34 * ( 20 ) = 86.8 m3 Table 6.5 Clarifier Sizing

Diameter (D)

2.5 m

High (H)

11 m

Retention time (TR)

20 minutes


4.34 m3/min

Volume of dorr clarifier needed

86.8 m3

48

6.6 Sizing of Heater (Unit - 6): The area of heater is depending on overall heat transfer coefficient "U" (W/m2.oC). The amount of "U" for organic compounds (sugar) with water is between (500 – 750) "U" = 700 W/m2.oC The amount of heat required in heater is calculated from the general equation of heat exchanger (6.2)

 =    ΔTlm =


ln

T ou t

, ΔTout = (100-92)= 8oC ,ΔTin=(130-88)= 42oC

T in

ΔTlm = (8 – 42) / ln (8/42) = 20.5 oC

Qheater = 3542618.875 kJ/h (Appendix B.6). Hence, the area of heater can be calculated now : A = (3542618.875*103/3600) / (700*20.5) = 68.56 m2 Table 6.6 Heater Sizing


700 W/m2.OC


20.5 OC

Amount of heat needed in heater (Q)

3542618.875 KJ/h

The area of heater (A)

68.56 m

49

6.7 Sizing of Sulfitation (Unit - 9): Assuming t= 1 h m = 43703.125 Kg/h and ρ= 1343.30 kg/m3



Volume required =  = 32.5 m3 6.8 Sizing of Crystallizer (Unit - 10): Assuming t= 1.5 h m = 32893.75 kg/h and ρ= 1586.2 kg/m3



volume required=  = 31.11 m3 6.9 Sizing of Cooler (Unit - 11): The area of cooler is depending on overall heat transfer coefficient "U" (W/m2.oC). The amount of "U" for organic compounds (sugar) with water is between (500 – 750) "U" = 650 W/m2.oC The amount of heat required in cooler is calculated from the general equation of heat exchanger (6.2)

 =    ΔTlm =


ln

T ou t

, ΔTout = (75-50)= 25oC ,ΔTin=(100-25)= 75oC

T in

ΔTlm = (25 – 75) / ln (25/75) = 45.5 oC

Qcooler = 1203015.6 w (Appendix B.10). 50

Hence, the area of cooler can be calculated now : A = (1203015.6) / (650*45.5) = 40.7 m2

Table 6.7 Cooler Sizing


650 W/m . C


45.5 C

Amount of heat needed in cooler (Q)

1203015.6 w

The area of cooler (A)

40.7 m

6.10 Sizing of Centrifuge (Unit -12):

Table 6.8 Centrifuge Sizing [11]

Model

TITAN 2400

Basket volume (m3)

1.4

Capacity (Kg)

2413

Length (m)

6.299

Diameter (m)

1.981

51

6.10 Sizing of Storage (Unit - 15):

Data of sugar : Mass flowrate of sugar (ms) = 300 ton/day = 300000 Kg/day Density of sugar (ρs) = 1586.2 Kg/m3

Volumetric flowrate of sugar (Vs) = ms/ρs = 189.13 m3/day = 1323.92 m3/week Assume : Length (L) = 12m , Diameter (D) = 3.5m

   ( =  D   = 11.4     =  =  

(6.4)[5]

52

Chapter 7. Water Recycling and Treatment 7.1

Overview:

Water recycling is used on four units in this project : washing ,diffusion tower , evaporation and cooling tower. The objectives of this operation is to reduce water consumption to reduce costs and preserve the natural resources .

Figure 7.1 (Diagram of Water Recycling and Treatment)

53

7.2

Washing Water Recycling:

In this plant ,we need one ton of water to clean two ton of beet of beet . Most of this water is supplied by various in-factory sources such as condensate, condenser water, and wastewater treatment system.

Figure 7.2 (Diagram of Washing Water Recycling)

Table 7.1 (Mass Balance) Balance)

Stream

Water(ton/day)

Beet(ton/day)

Mud(ton/day)

1

0

3554

396

2

2370

0

0

3

0

3554

0

4

7900

0

396

5

5530

0

0

6

2370

0

396

7

1659

0

0

8

711

0

396

54

7.3

Diffusion Tower Water Recycling:

The pulp exiting exiting from diffuser has has 90% water water and 10% solid . We need to compress the pulp to extract the water in it. After that drain the pulp to be sold as food for animals. 10

9

12

11

Figure 7.3(Diagram of diffusion tower water recycling)

Table 7.2 (Mass Balance) Balance)

7.4

Stream

Water(ton/day)

Pulp(ton/day)

9

639.72

71.08

10

511.78

0

11

127.94

71.08

12

0

71.08

Evaporator Water Recycling To Diffusion Tower:

Diffusion Tower

13 Evaporator

Figure 7.4 (Diagram of Evaporator Water Recycling to Diffusion Tower)

Table 7.3 (Mass Balance)

Stream

Water(ton/day)

13

3845.875

55

7.5 Cooling Water Recycling From Cooling Tower:

Figure 7.5 (Diagram of Cooling Water Recycling From Cooling Tower)

Mass balance:Mass flowrate of water (W) : Cp,water at Tav= (40+25)/2= 32.5oC ,Cp,water = 4.18 (kJ/kg.oC) ΔTwater = ( T2- T1) = (40 - 25) = 15oC

Q = mwater (Cpwater ) ( ΔTwater ) = mwater (4.18)(15) mwater =

14171.4 (kg/h) = 340.11 (ton/day)

Mass flowrate of air (A) : Cp,air at Tav = (38+20)/2= 29 oC, Cp,air = 1.005 kJ/Kg.oC Δtair = ( t2- t1) = (38 - 20) = 18oC

Q = mair (Cpair ) (ΔTair ) = mwater (1.005)(18) mair =

49118.125 (kg/h)

56

7.6

Water Treatment :

The major aim of wastewater treatment is to remove as much of the suspended solids as possible before the remaining water, called effluent, is discharged back to the environment. There are many types of water treatment. Chlorination is the most famous type because it is easily available and not expensive.



Chlorination :

Chlorination is the process of adding the element chlorine to water as a method of water purification to make it fit for human consumption as drinking water. Water which has been treated with chlorine is effective in preventing the spread of waterborne diseases. The chlorination of public drinking supplies was originally met with resistance, as people were concerned about the health effects of the practice. The use of chlorine has greatly reduced the prevalence of waterborne disease as it is effective against almost all bacteria and viruses, as well as amoeba. From flow sheet , we need to treat the amount of water exit from filter in washing process by adding chlorine (Cl2) to get useable pure water.

57

Chapter 8.HAZOP and Environment 8.1

HAZOP

A Hazard and Operability (HAZOP) study is a structured and systematic examination of a planned or existing process or operation in order to identify and evaluate problems that may represent risks to personnel or equipment, or prevent efficient operation. The HAZOP technique was initially developed to analyze chemical process systems, but has later been extended to other types of systems and also to complex operations. A HAZOP is a qualitative technique based on guide-words and is carried out by a multi disciplinary team (HAZOP team) during a set of meetings. The hazard is any operation that could possibly cause a catastrophic release of toxic, flammable or explosive chemicals or any action that could result in injury to personnel. The operability is any operation inside the design envelope that would cause a shutdown that could possibly lead to a violation of environmental, health or safety regulations or negatively impact profitability. HAZOP study will be covered the following equipment : - Reactor - Cooling Tower - Evaporator - Dryer

58

8.1.1 Reactor

Table 8.1 HAZOP of Reactor Unit

Guide Word No

Reverse

Deviation Causes

No flow cooling

Reverse Flow

More of

More cooling flow

As well as

Reactor product in Coils

Reactor

Other than

-Another material besides cooling water

-Cooling water valve Malfunction - Failure of water source resulting in backward flow - Control valve failure, operator fails to take action on alarm - More pressure in reactor

Consequences Action - Temperature -Install high increase in temperature reactor alarm (TAH)

- Less cooling, possible runaway reaction

- Too much cooling, reactor cool

- Off-spec product

-May be cooling -Water inefffective source and Contaminated effect on the reaction

- Install check valve

Instruct operators on Procedures - Check maintenance procedures and schedules -If less cooling, TAH will detect. If detected, isolate water source. Back up water source?

59

8.1.2 Cooling Tower

Unit

Guide Word No

Table 8.2 HAZOP of Cooling Tower Deviation Causes Consequences No flow - Failure of - Process fluid cooling inlet cooling temperature is water not valve to open lowered accordingly

Less

Less cooling flow

- Pipe leakage - Process fluid temperature too low

More

More cooling flow

- Reverse process fluid Flow

- Output of Process fluid temperature too low

Reverse

Reverse cooling Flow

- Failure of process fluid inlet Valve

- Product off set

Other than

Other material -Outlet than water Contamination temperature in cooling too low water

Cooling tower

Action - Install Temperature indicator before and after the process fluid line Install TAH - Installation of flow meter

- Install Temperature indicator before and after process fluid line Install TAL - Install check valve (whether it is crucial have to check) - Proper maintenance and operator alert

60

8.1.3 Evaporator

Unit

Guide Word No

Table 8.3 HAZOP of Evaporator Deviation Causes Consequences No steam flow - Failure of - Process fluid inlet steam temperature is not valve to open increased accordingly

Less

Less steam

- Pipe leakage

- Process fluid temperature low

Evaporator More

More steam flow

- Reverse process fluid Flow

- Output of Process fluid temperature too high

Reverse process fluid Flow

- Failure of process fluid inlet Valve

- Product off set

Reverse

Action - Install Temperature indicator before and after the process fluid line Install TAL - Installation of flow meter - Install Temperature indicator before and after process fluid line Install TAH - Install check valve

61

8.1.4 Dryer

Table 8.4 HAZOP of Dryer

Unit

Rotary drum dryer

Guide Word No

Deviation No hot Air flow

Causes - Failure of inlet hot air valve to open

Consequences - Process fluid temperature is not Increased accordingly

Less

Less hot Air

- Pipe leakage

- Process fluid temperature low

More

More hot Air flow

- Reverse process fluid flow

- Output of Process fluid temperature too high

Reverse

Reverse process fluid Flow

- Failure of process fluid inlet valve

- Product off set

Action -Install Temperature indicator Before and after the process fluid line Install TAL - Installation of flow meter -Install Temperature indicator before and after process fluid line Install TAH - Install check valve

62

8.2

Environmental

"We are dependent on water and air in every minute of our life. Thus , it is the responsibility of every one of us to not degrade the natural environment with solid, liquid, and air pollutants. The systems for treatment of wastewater and solid wastes for reuse and to release safe enivroment . The beet-sugar factory produces more waste products than sugar, pulp, and molasses combined. If beet-sugar producers are not diligent in caring and disposing of waste products, the wastes can pollute the natural environment significantly. It makes good business sense for producers to reduce pollution and comply with all legally required actions and limitations with regard to the creation, storage, treatment, and disposal of pollutants, wastes, and hazardous compounds. Not doing so invites costly enforcement actions by the governments and lawsuits by citizens. If comprehensive pollution reduction is approached correctly, sugar producers not only be environmentally safe but also can generate considerable cost savings."[1]

8.2.1 Pollution Control Strategies

"Managers at each facility have to determine what combination of strategies makes sense for their situation. Weather, soil types, ground and surface water, facility age, current equipment,local population, local regulatory interests, costs and return on investment are some of the factors that managers must balance when selecting and implementing pollution control strategies. Pollution control strategies fall into one of the following four categories: ■ Prevention ■ Treatment ■ Measured release ■ Storage

After management has selected and implemented a strategy, plant personnel must operate and maintain the systems to economically perform the required

63

functions. One way to control pollution is to not create polluting compounds or conditions through energy efficiency, better controls on operations, changes in processes, and reduced wasting. Increasing energy efficiency reduces pollutants associated with energy generation. Facilities can reduce the amount of waste water through changes in beet transport and washing."[1]

8.2.2 Emissions to Air

"Air emissions in sugar manufacturing are primarily related to particulate matter generated from bagasse-fired steam boilers,dust from unpaved access roads and areas, and sugar drying or packing activities. In addition, odor emissions are generated from beet processing activities and storage facilities. Beet factory juice clarification produces a sweet odor, which can be irritating. Inadequate cleaning of the raw material may result in fermented juice, which will also create a foul smell."[12]

64

8.2.3 Safety :

Table 8.5 MSDS for Sugar Factory * - First aid measures :

Swallowed: Give water to drink. Eye: Flush thoroughly with copious

amounts

of

symptoms

running

persist,

water.

seek

If

medical

attention. Skin: Wash thoroughly with soap and

water. Inhaled: Remove to fresh air. Advice

to

Doctor:

symptomatically.

Treat

People

with

diabetes may need stabilization. - Flammability :

Low, product will burn in surrounding fire situation.

- Extinguishing Media :

Water, dry chemical, carbon dioxide, foam.

- Handling :

Material can ferment if excessive moisture contamination is allowed. Fermentation

can

yield

carbon

dioxide with possible traces of ethanol or volatile fatty acids (e.g. acetic, propionic, lactic, or butyric) and if exposed to a spark or flame may result

in

an

explosion.

These

conditions should be avoided. If maintenance of tank requires entry by personnel,

confined

space

precautions should be complied with. Insufficient oxygen may be present in vessels containing the product due to the generation of carbon monoxide 65

during fermentation. - Storage:

This product should be stored in its factory packaging in a dry area.

- Personal protection:

Eye Protection : Ventilated non-

fogging goggles should be worn if dust is generated. Skin Protection : Direct skin contact

should be avoided by wearing long sleeved shirts and long trousers, a cap or hat, and gloves Work clothes should be washed regularly. - Chemical Stability:

Stable.

- Toxicity Data :

Non-toxic .

- Health Effects:

Swallowed : No health effects under

normal conditions of industrial use, but ingestion may destabilise people with diabetes. Eye: Irritating to the eyes and may

cause watering and redness . Skin: Skin contact may result in mild

skin irritation .

*

MSDS : Material Safety Data Sheet for (white sugar – pulp – molasses)

66

Chapter 9. Control Loop 9.1

Overview :

Control Unit is an important process to observe the conduct of operations in the factory. There are three forms of control operations: o

Flow control (FC).

o

Temperature control (TC).

o

Concentration control (CC).[6]

Figure 9.1 Control Flow Sheet

67

9.2 Diffusion Tower Control :

Figure 9.2 Diffusion Tower Control

9.3

Mixer Control :

Figure 9.3

Mixer Control

68

9.4

Heater(1) Control :

Figure 9.4 Heater1 Control

9.5

Liming Control :

Figure 9.5 Liming Control

69

9.6

Clarifier Control :

Figure 9.6 Clarifier Control

9.7

Heater(2) Control :

Figure 9.7 Heater2 Control

9.8

Filter Control :

Figure 9.8 Filter Control

70

9.9

Evaporator Control :

Figure 9.9 Evaporator Control

9.10 Sulfitation Control :

Figure 9.10 Sulfitation Control

9.11

Crystallizer Control :

Figure 9.11 Crystallizer Control

71

9.12

Cooler Control :

Figure 9.12 Cooler Control

9.13

Centrifuge Control :

Figure 9.13 Centrifuge Control

9.14

Dryer Control :

Figure 9.14 Dryer Control

72

9.15

Cooling Tower Control :

Figure 9.15 Cooling Tower Control

73

Table 9.1 : Control Overview

Equipment

Manipulated

Controlled

Action

Variables

Variables

Diffusion Tower

St.14 Flow rate

Flow rate

Control Valve ( B1 )

Mixer

St.1 Flow rate

Flow rate


Liming

St.7 Flow rate

Flow rate


St.8 Flow rate

Exit composition

Control Valve ( B5)

Clarifier

St.8 Flow rate

Flow rate

Control Valve ( B6)

Filter

St.13 Flow rate

Flow rate


Evaporator

St.13 Flow rate

Flow rate


St.15 Flow rate

Temperature

Control Valve ( B9)

Sulfitation

St.16 Flow rate

Flow rate

Control Valve ( B10)

Crystallizer

St.16 Flow rate

Flow rate


St.18 Flow rate

Temperature


Centrifuge

St.20 Flow rate

Flow rate


Dryer

St.21 Flow rate

Temperature


Cooling Tower

St.22Flow rate

Temperature


Heater (1)

St.6 Flow rate

Temperature


Heater (2)

St.11Flow rate

Temperature


Cooler

St.19 Flow rate

Temperature


74

Chapter 10. Cost Analysis 10.1 Overview

The cost is a rough estimate to evaluate a preliminary cost for the equipment of the sugar production to make quick decision to the project evaluation . The cost of equipment can be found by the following formula : [7]

(10.1)

10.2 Methods for estimating capital investment

Various methods can be employed for estimated capital investment. The choice of any one methods depends on the amount of available information and the accuracy desired. Seven methods are listed below : 1. Detailed-Item Estimate Method. 2. Unit-Cost Estimate Method. 3. Percentage of Delivered Equipment Cost Method. 4. Lang Factor for Approximation of Capital Investment Method. 5. Power Factor Applied to Plant/Capacity Ratio Method. 6. Investment Cost per Unit of Capacity Method. 7. Turnover Ratio Method.[7]

10.3 Equipment cost purchased in 2007 [14]

Diffusion Tower = $ 77200 Mixer 1 = $ 681800 Heater 1 = $ 61,100 Liming tank = $ 512,600 75

Reactor 2 = $ 627,900 Clarifier = $ 389,900 Heater = $ 32,200 Filter = $ 380,800 1st Evaporator = $ 3,253,000 2nd Evaporator = $ 2,489,700 3rd Evaporator = $ 2,690,800 Crystallizer = $ 152,300 Sulfitation Tank = $ 36300 Cooler 1 = $ 205,100 Centrifuge = $ 203,500 Dryer = $ 519,400 Cooler 2 = $ 79,300 Storages = $ 321,600 Total purchased equipment = $ 12,699,800 Cost index at 2007 = 525 Cost index at 2012 = 580

Cost of equipment in KSA in 2012 =

      (2007           (10.2)

[7]

Location factor = 1.4 "because the plant location in KSA"

10.4 Equipments In KSA at 2012 :

Diffusion Tower : Type

Diffuser

Pressure

2

atm

Volume

310.65

m3

Material

Carbon steel

Cost of diffuser at 2012 in KSA = 77200*(580/525)*1.4 = $ 119402

76

Static Mixer : Pressure

2

Atm

Volume

85.95

m

Material

Carbon steel

Cost of mixer at 2012 in KSA = 681800*(580/525)*1.4 = $ 1,054,517 Heater 1 : Type

Shell and tube floating head

Pressure

2

Atm

Area

175.5

m2

Material

Carbon steel

Cost of heater at 2012 in KSA = 61100*(580/525)*1.4 = $ 94,501 Reactor 1 : Type

Continuous stirred tank reactor

Pressure

2

atm

Volume

84.46

m

Material

Stainless steel

Cost of reactor at 2012 in KSA = 627900*(580/525)*1.4 = $ 971,152

Liming Tank : Pressure

2

atm

Volume

64

m

Material

Carbon steel

Cost of tank at 2012 in KSA = 512600*(580/525)*1.4 = $ 792,821 Clarifier : Type

Dorr rake

Pressure

2

atm

Diameter

2.5

m

Material

Carbon steel

Cost of clarifier at 2012 in KSA = 389900*(580/525)*1.4 = $ 603,045

77

Heater 2 : Type

Shell and tube floating head

Pressure

2

atm

Area

68.56

m2

Material

Carbon steel

Cost of heater at 2012 in KSA = 32200*(580/525)*1.4 = $ 49,803 Filter : Type

Rotary drum (continuous)

Pressure

1

atm

Area

67.7

m

Material

Carbon steel

Cost of filter at 2012 in KSA = 380800*(580/525)*1.4 = $ 588,971

Evaporator 1 : Type

Robert (vertical tube)

Pressure

1

atm

Area

947.79

m

Material

Stainless steel

Cost of 1st effect evaporator at 2012 in KSA = 3253000*(580/525)*1.4 = $ 5,031,307

Evaporator 2 : Type


Pressure

1

atm

Area

640.36

m2

Material

Stainless steel

Cost of 2nd effect evaporator at 2012 in KSA = 2489700*(580/525)*1.4 = $ 3,850,736

78

Evaporator 3 : Type


Pressure

1

atm

Area

717.6

m2

Material

Stainless steel

Cost of 3rd effect evaporator at 2012 in KSA = 2690000*(580/525)*1.4 = $ 4,160,533 Crystallizer : Type

Batch

Pressure

1

atm

Volume

31.11

m

Material

Carbon steel

Cost of crystallizer at 2012 in KSA = 152300*(580/525)*1.4 = $ 235,557

Sulfitation Tank : Pressure

1

atm

Volume

32.4

m3

Material

Carbon steel

Cost of tank at 2012 in KSA = 36300*(580/525)*1.4 = $ 56144 Cooler 1 : Type

Cooling tower , natural draft

Pressure

1

atm

Area

293.7

m

Material

Carbon steel

Cost of cooler at 2012 in KSA = 205100*(580/525)*1.4 = $ 317,221

79

Centrifuge : Type

Vertical , Batch

Pressure

1

atm

Diameter

1.981

m

Material

Carbon steel

Cost of centrifuge at 2012 in KSA = 203500*(580/525)*1.4 = $ 314,747 Dryer : Type Area

Rotary drum m2

89.87

Material

Stainless steel

Cost of dryer at 2012 in KSA = 519400*(580/525)*1.4 = $ 803,339 Cooler 2 : Type

Cooling tower , natural draft

Pressure

1

atm

Area

17.55

m

Material

Carbon steel

Cost of cooler at 2012 in KSA = 79300*(580/525)*1.4 = $ 122,651 Storages : Type

Silos

Number

12

Pressure

1

atm

Volume

115.45

m3/silos

Material

Carbon steel

Cost of 12 storage at 2012 in KSA = 321600*(580/525)*1.4 = $ 497,408

TOTAL PURCHASED EQUIPMENTS AT 2012 IN KSA = $ 19,544,453

80

10.5

Total Capital Investment (TCI) :

Total Capital Investment divided into Fixed Capital Investment (FCI) and Working Capital Investment (WCI) as follows in equation (10.3) : [7]

 =   

(10.3)

Where : Fixed Capital Investment = Direct Costs + Indirect Costs

(10.4)

Direct Costs = $ 59,024,247 Indirect Costs = $ 24,626,011 Fixed Capital Investment (FCI) = $ 83,650,258 Working Capital Investment = 75% of Purchased equipments cost Working Capital Investment (WCI) = (0.75 * 19,544,453) = $ 14,658,340 Total Capital Investment (TCI) = FCI + WCI TCI = $ 83,650,258 + $ 14,658,340 = $ 98,308,598

81

Table 10.1 : The Total Capital Investment In Sugar Factory [7]

Item

Cost Capital %

Cost ($)

Direct Costs

Purchased equipments cost

100

19,544,453

Purchased equipments installation

39

7,622,337

Instrumentation and controls

26

5,081,558

Piping (installed)

31

6,058,780

Electrical system (installed)

10

1,954,445

Buildings (including service)

29

5,667,891

Yard improvement

12

2,345,334

Service facilities

55

10,749,449

302

59,024,247

Engineering and supervision

32

6,254,225

Construction expenses

34

6,645,114

Contractor's fee

19

3,713,446

Contingency

37

7,231,448

Legal expenses

4

781,778

Total indirect costs

126

24,626,011

FCI $

428

83,650,258

Working capital (15% T.C.I)

75

14,658,340

Total Capital Investment

503

98,308,598

Total direct plant costs Indirect Costs

82

10.6

Total Production Cost (TPC) :

Total production cost is divided into the manufacturing cost (MC) and general expenses (GE).

 =   

(10.5)

Annual production = (300 ton/day) * (180 day/year) = 54,000 ton/year Our plant will be operating for 6 months : Table 10.2 : Raw Material Cost

Raw Material

Cost ($/ton)

ton/year

$/year

41

711,000

29,151,786

0.267

1,422,000

379,200

CO2

11

24,562

270,182

SO2

2.12

180

381.6

80

24,080

1,926,400

Beet Water

Lime (CaCO3)

Total Raw Material Cost (TRMC)

31,727,164

83

Table 10.3 : The Total Production Cost [7]

Item

Cost capital %

Cost ($)

MANUFACTURE COST Direct Production Cost (DPC)

Raw Material

31,727,164

31,727,164

Operating Labor (OL)

10% TPC

0.1 TPC

Direct supervisory and Electrical Labor

10% OL

0.01 TPC

Utilities

10% TPC

0.1 TPC

Maintenance and Repairs

4% FCI

3,346,010

Operation Supplies

0.5% FCI

418,251

Laboratory Supplies

10% OL

0.01 TPC

Patents and Royalties

3% TPC

0.03 TPC

Fixed Charges (FC)

11,711,036

Depreciation

10% FCI

8,365,026

Local Taxes

---

---

Insurance

0.4% FCI

3,346,010

Plant Overhead Cost (POC)

6% TPC

0.06 TPC

GENERAL EXPENSES

Administration Cost

2 % TPC

0.02 TPC

Distribution and Marketing Cost

5% TPC

0.05 TPC

Research and Development Costs

5% TPC

0.05 TPC

TOTAL PRODUCTION COST (TPC)

47,202,461 + 0.43 TPC

From table above : TPC = 47,202,461 + 0.43 TPC  TPC = (47,202,461) / (0.57) = $ 82,811,335

84

Table 10.4 : The calculations of Total Production Cost

Item

Cost ($)

MANUFACTURE COST = 72,873,974 Direct Production Cost (DPC)

56,194,259

Raw Material

31,727,164

31,727,164

Operating Labor (OL)

10% TPC

8,281,134

Direct supervisory and Clerical Labor

10% OL

828,113

Utilities

10% TPC

8,281,134

Maintenance and Repairs

4% FCI

3,346,010

Operation Supplies

0.5% FCI

418,251

Laboratory Supplies

10% OL

828,113

Patents and Royalties

3% TPC

2,484,340

Fixed Charges (FC)

11,711,036

Depreciation

10% FCI

8,365,026

Local Taxes

---

---

Insurance

0.4% FCI

3,346,010

Plant Overhead Cost (POC)

4,968,680

GENERAL EXPENSES = 9,937,361

Administration Cost

2 % TPC

1,656,227

Distribution and Marketing Cost

5% TPC

4,140,567

Research and Development Costs

5% TPC

4,140,567

TOTAL PRODUCTION COST (TPC)

82,811,335

85

10.7 Profitability Analysis :

Rate of Return can be calculated in equation (10.6) as follows : [7]

   ( =

  –   100 (10.6)   

Sales = [White Sugar] + [Molasses] + [Pulps] White Sugar = 54,000,000 (Kg/year) * 1 ($/Kg) = 54,000,000 $/year Molasses = 8,526,600 (Kg/year) * 10 ($/Kg) = 85,266,000 $/year Pulps = 12,794,400 (Kg/year) * 0.19 ($/Kg) = 2,430,936 $/year Sales = 54,000,000 + 85,266,000 + 2,430,936 = $ 141,696,936/year Net profit = Sales – Total Production Cost = 141,696,936 – 82,811,335 = $ 58,885,601/year

   ( =

  100 = 59.89 %

Since the plant operates for 6 months per year : 59.89 / 2 = 29.95 %

10.8

Payback Period (PBP)

Payback Period =

     

(10.7)[7]

Annual Cash Flow = Annual Profits + Depreciation Depreciation = $ 8,365,026 Annual Cash Flow = 58,885,601 + 8,365,026 Annual Cash Flow = $ 67,250,627/year

  

Payback Period =     Payback Period =

(10.8)[7]

  = 1.46 

The plant operates for (6 months)/year

.

Payback Period =  = (

2.92  86

Table 10.5 : Summary for Economics Calculations

1

Total capital Investment

$ 98,308,598

2

Fixed Capital Investment

$ 83,650,285

3

Working Capital Investment

$ 14,658,340

4

Total Production Costs

$ 82,811,335

5

Rate of Return

29.95 %

6

Payback Period

2.92 Year

87

CHAPTER 11. Conclusion This project is related with production of sugar from beets using the carbonation process . The production rate was fixed to 300 ton/day of sugar. The main product consists of 150 ton of normal sugar and 150 ton of specially sugar (powder). The pulp of the slicing beet after diffusing will be sold as byproduct used as animals food. The overall and individual material and energy balances have been performed . Furthermore, the HAZOP, safety and environmental analysis has been conducted for the major equipments. On the other hand, the control loop has been carried out for the overall flowsheet and individual major equipments where both the control and manipulated variables have been specified. The treatment and recycling of the large amount of water requested in different parts of the process (especially in the washing and diffusion steps) have reduced the total cost of the plant. Based on this study and since the raw materials are not very expensive, the cost estimation confirms the profitability of the carbonation process.

88

REFERENCES :

Books : [1] Mosen Asadi (2007). “Beet-Sugar Handbook” .

[2] Felder and Rousseau (2005). "Elementary Principles of Chemical Process". 3rd edition. [3] Fogler and Gurmen (2011). "Essentials of Chemical Reaction Engineering". 4th edition. [4] Christie John Geankoplis (2008). "Transport Process and Separation Process Principles". 4th edition. [5] McCabe, Smith and Harriott (2005). "Unit Operation of Chemical Engineering". 7th edition. [6] Stephanopoulos, G (2003). "Chemical Process Control: An Introduction to Theory and Practice". [7] Peters and Timmerhaus (1991). "Plant Design and Economics for Chemical Engineering". 4th edition.

89

Websites :

[8] http://www.food-info.net/uk/products/sugar/history.htm [9]http://www.agmrc.org/commodities__products/grains__oilseeds/sugarcane

profile.cfm [10] (http://blog.ub.ac.id/spiritgumilang/2012/10/05/single-multiple-effectevaporator/) [11] www.westernstates.com/centrifuge [12] http://ar.scribd.com/doc/6749495/Sugar-Safety

[13] www.klmtechgroup.com [14] www.matche.com/equipment

90

APPENDICES APPENDIX A : MASS BALANCE A.1 Diffusion balance

Water

14 (C12H22O12) Juice diff.

beet

XDS = 0.15 Xw = 0.85

Diffusion 5

1 4

Pulp X p= 0.2 onbeet

Overall M.B. : m1+ m14= m4 + m5

 m1+

3845.87= m4 + 6689.05

m4=0.2m1  m1+3845.87=

0.2m1 + 6689.05

 0.8m1=6689.05-3845.87  m1=3554  m4=

ton/day = 148083.33 kg/h

(0.2)(3554)=710.8 ton/day = 29616.67 kg/h

Component M.B. : STREAM 5 : m5DS = 0.15 (6689.05) = 1003.36 ton/day =41806.67 kg/h m5S = 1003.36 (0.88) = 883 ton/day = 36791.67 kg/h m5NS = 1003.36 (0.12) = 120.36 ton/day = 5015 kg/h m5w = 0.85 (6689.05) = 5685.7 ton/day = 236904.2 kg/h

91

A.2 Mixer balance : Juice tower

(C12H22O12)

3

XDS = 0.15

Xw = 0.85

Mixer Beet

Beet

1

2

+ 0.8 juice (on

5

beet)

Juice diff.

(C12H22O12) XDS = 0.15 Xw = 0.85

DS= Dry substance (88% S , 12% NS) W = Water (Purity = 88 %) Overall M.B. : m1+ m3 = m2 + m5  3554+ m3 = m2 + 6689.05  m2=

m1+ 0.8m5

 m3=m2+m5-m1=

 m2=3554+(0.8)(3554)=

6397.2 ton/day = 266550 kg/h

6397.2+ 6689.05- 3554= 9532.25 ton/day =397177.1 kg/h

92

At (88%) purity STREAM (3) : m3 = 9532.25 ton/day =397177.1 kg/h m3DS = 9532.25 (0.15) = 1429.84 ton/day = 59576.67 kg/h (88% S, 12% NS) m3S = 1429.84 (0.88) = 1258.26 ton/day = 52427.5 kg/h m3NS = 1429.84 (0.12) = 171.58 ton/day = 7149.17 kg/h STREAM (5) : m5DS = 6689.05 (0.15) = 1003.36 ton/day = 41806.67 kg/h (88% S, 12% NS) m5S = 1003.36 (0.88) = 883 ton/day = 36791.67 kg/h m5NS = 1003.36 (0.12) = 120.36 ton/day = 5015 kg/h m5W = 0.85 (6689.05) = 5685.7 ton/day = 236904.2 kg/h STREAM (2) m2DS = 6397.2 (0.15) = 959.58 ton/day = 39982.5 kg/h m2W = 6397.2(0.85) = 5437.62 ton/day = 226567.5kg/h

93

A.3 Diffuser balance :

Water Beet +

0.8 juice

14

2

Pulp

(on beet)

4 diffuser

Juice tower (C12H22O12)

Xp= 0.2 onbeet

3

XDS = 0.15 Xw = 0.85

Overall M.B. : STREAM (2) m2DS = 6397.2 (0.15) = 959.58 ton/day = 39982.5 kg/h m2W = 6397.2(0.85) = 5437.62 ton/day = 226567.5kg/h STREAM (3) : m3 = 9532.25 ton/day =397177.1 kg/h m3DS = 9532.25 (0.15) = 1429.84 ton/day = 59576.67 kg/h (88% S, 12% NS) m3S = 1429.84 (0.88) = 1258.26 ton/day = 52427.5 kg/h m3NS = 1429.84 (0.12) = 171.58 ton/day = 7149.17 kg/h STREAM(4): m4=710.8 ton/day = 29616.67 kg/h STREAM(14): m14=3845.87 ton/day =160244.58 kg/h

94

A.4 Liming balance : MOL

(Ca(OH)2) Xl = 0.02

7 6

(C12H22O12) XDS = 0.15 Xw = 0.85

8 Liming

(C12H22O12) XDS = 0.15 Xw = 0.85

(Purity = 88 %) Overall M.B. : m6+ m7 = m8  m6+ m7 = 6822.83  m6 +0.02 m6 = 6822.83  1.02m6 =  m6 =

6822.83

6689.05 ton/day = 278710.42 kg/h

m6DS = 6689.05 (0.15) =1003.36 ton/day = 41806.67 kg/h(88% S,12% NS) m6S = 1003.36 (0.88) = 883 ton/day = 36791.67 kg/h m6NS = 1003.36 (0.12) = 120.36 ton/day = 5015 kg/h m6W = 0.85 (6689.05) = 5685.7 ton/day = 236904.2 kg/h At (STREAM 7) Hence, the amount of milk of lime can be calculated as follows :

m7 = m8 – m6

= 6822.83 – 6689.05 = 133.78 ton/day = 5574.2 kg/h

95

A.5 Clarifier balance :

8

(C12H22O12) XDS = 0.15 Xw = 0.85

CLARIFIER

10

(C12H22O12) XDS = 0.15 Xw = 0.85

9 Slurry (Mud) XDS = 0.7 XW = 0.3

DS= Dry substance (88% S , 12% NS) W = Water (Purity = 88 % and m9 = 25 % m8 and 70 % of m9 DS ) Overall M.B. : m8 = m9 + m10

 m8 =

m9 + 5117.12

m8 = 0.25 m8 + 5117.12  m8 =  m9 =

 0.75

m8 = 5117.2

6822.83 ton/day = 284284.5833 kg/h m8 – m10 = 6822.83 – 5117.12 = 1705.71 ton/day = 71071.25 kg/h

Component M.B. : STREAM 8 : m8DS = 0.15 (6822.83) = 1023.42 ton/day = 42642.5 kg/h m8S = 0.88 (1023.42) = 900.6096 ton/day = 37525.4 kg/h m8NS = 0.12 (1023.42) = 122.8104 ton/day = 5117.1 kg/h m8w = 0.85 (6822.83) = 5799.4055 ton/day = 241641.8958 kg/h STREAM 9 : m9DS = 0.7 (1705.71) = 1193.997 ton/day = 49749.875 kg/h m9w = 0.3 (1705.71) = 511.713 ton/day = 21321.375 kg/h

96

A.6 Filter balance :

(C12H22O12) XDS = 0.15 Xw = 0.85

11

FILTER

13

(C12H22O12) XDS = 0.15 Xw = 0.85

12 REJECT XDS = 0.15 Xw = 0.85

DS= Dry substance (92% S , 8% NS) W = Water (Purity = 92 %) Overall M.B. : m11 = m12 + m13  m11 = m12 + 4894.75  m12 = m11 – 4894.75 At (92%) purity (STREAM 13) : m13 = 4894.75 ton/day = 203947.9167 kg/h m13DS = 4894.75 (0.15) = 734.2125 ton/day = 30592.1875 kg/h (92% S, 8% NS) m13S = 734.2125 (0.92) = 675.4755 ton/day = 28144.8125 kg/h m13NS = 734.2125 (0.08) = 58.737 ton/day = 2447.375 kg/h In filtration , the inlet has (88%) purity , the filtrate has (92%) purity , so we can find the mass flow rate of the inlet (STREAM 11) as follows : m11T = m11DS + m11W m11DS = (675.4755/0.88) = 767.585 ton/day = 31982.708 kg/h (88% S, 12%NS)

97

m11S = 767.585 (0.88) = 675.474 ton/day = 28144.7833 kg/h m11NS = 767.585 (0.12) = 92.11 ton/day = 3837.925 kg/h m11T = (767.585/0.15) = 5117.2 ton/day = 213218.1 kg/h m11w = m11T – m11DS = 4349.6 ton/day = 181233.95 kg/h STREAM 12 : m12 = m11 – 4894.75  m12 = 222.45 ton/day = 9268.75 kg/h m12w = 0.85 (222.45) = 189.0825 ton/day = 7878.4375 kg/h m12DS = m12NS = 0.15 (222.45) = 33.368 ton/day = 1390.3125 kg/h

98

A.7 Evaporator balance :

VAPOR XDS = 0.0 Xw = 1.0

14

(C12H22O12) XDS = 0.15 Xw = 0.85

13

15 Evaporator

(C12H22O12) XDS = 0.7 Xw = 0.3

(Purity = 92 %) Overall M.B. : m13 = m14 + m15  m13 = m14 + 1048.875  m14 = m13 – 1048.875 Component M.B. : For water : (STREAM 13) 0.85 m13 = (m13 – 1048.875) + 0.3(1048.875) -0.15 m13 = -734.2125  m13 =

4894.75 ton/day = 203947.9167 kg/h

m13DS = 4894.75 (0.15) = 734.2125 ton/day = 30592.1875 kg/h (92% S, 8% NS) m13S = 734.2125 (0.92) = 675.4755 ton/day = 28144.8125 kg/h m13NS = 734.2125 (0.08) = 58.737 ton/day = 2447.375 kg/h m13W = 0.85 (4894.75) = 4160.5375 ton/day = 173355.7292 kg/h At (STREAM 14) XDS = 0 and XW = 1.0 Hence, the amount of water vaporized during evaporation can be calculated as follows : m14 = m13 – m15 = 4894.75 – 1048.875 = 3845.875 ton/day = 160244.7917 kg/h

99

A.8 Sulfitation balance :

SO2

(C12H22O12) XDS = 0.7 Xw = 0.3

15

16

SULFITATION

(C12H22O12) XDS = 0.7 Xw = 03

(Purity = 92 %) Overall M.B. : m15= m16  1048.875 ton/day (43703.125 Kg/h) Component M.B. : (STREAM 15 = STREAM 16) 0.7 m16 = m16DS = m15DS = 43703.125(0.7) = 30592.1875 Kg/h m15S = m16S = 30592.1875 (0.92) = 28144.8125 kg/h m15NS = m16NS = 30592.1875 (0.08) = 2447.375 kg/h 0.3m16 = m16W = m15W = 0.3 (43703.125) = 13110.9375 kg/h

100

A.9 Crystallizer balance :

17

(C12H22O12) XDS = 0.7 Xw = 0.3

H2O

16

18

CRYSTALLIZER

(C12H22O12) XDS = 0.93 Xw = 0.07

(Purity = 92 %) Overall M.B. : m16 = m17 + m18  m16 = m17 + 798.45 ton/ day (32893.75 Kg/hr) m17 = m16 – 32893.75 Component M.B. : ( STREAM 16) mT = 43703.125= m16DS = 43703.125(0.7) = 30592.1875 Kg/h m16S = 30592.1875 (0.92) = 28144.8125 kg/h m16NS = 30592.1875 (0.08) = 2447.375 kg/h m16W = 0.3 (43703.125) = 13110.9375 kg/h ( STREAM 18) m18= 32893.75 Kg/hr m18DS = m16DS = 30592.1875 Kg/h m18S = m16S = 28144.8125 kg m18NS = m16NS = 2447.375 kg/h m18W = m18 - m18DS = 2301.5625 Kg/hr m17 (Only water): m17W= 43703.125 - 32893.75 = 10809.375 Kg/hr

101

A.10 Cooler balance :

(C12H22O12) XDS = 0.93 Xw = 0.07

(C12H22O12) XDS = 0.93 Xw = 0.07

COOLER

m18=m19 = 32893.75 Kg/hr A.11 Centrifuge balance :

(C12H22O12) XDS = 0.93 Xw = 0.07

19

20

CENTRIFUGE

Other processes XDS = 0.94 Xw = 0.06

21

(C12H22O12) XDS = 0.95

XW = 0.05

DS= Dry substance (92% S , 8% NS) W = Water (Purity = 92 % and m19 = 60 % m18 and m20 = 40 % of m18) Overall M.B. : m19 = m20 + m21

 m19 =

m20 + 315.78

m19 = 0.6m19 + 315.78  0.4m19 = 315.78  m19 =

789.45 ton/day = 32893.75 Kg/h

 m20 =

m19 - m21 = 789.45 – 315.78 = 473.7 ton/day = 19737.5 Kg/h 102

Component M.B. : STREAM 19 : m19DS = 0.93 (789.45) = 734.2 ton/day = 30591.7 kg/h m19S = 0.92 (734.2) = 675.46 ton/day = 28144.33 kg/h m19NS = 0.08 (734.2) = 58.736 ton/day = 2447.33 kg/h m19w = 0.07 (789.45) = 55.26 ton/day = 2302.56 kg/h STREAM 20 : m20DS = 0.94 (473.7) = 445.278 ton/day = 18553.25 kg/h m20w = 0.06 (473.7) = 28.422 ton/day = 1184.25 kg/h STREAM 21 : m21DS = 0.95 (315.78) = 300 ton/day = 12500 kg/h m21W = 0.05 (315.78) = 15.78 ton/day = 657.5 kg/h

103

A.12 Dryer balance :

(C12H22O12) XDS = 0.95 Xw = 0.05

21

DRYER

23

(C12H22O12) XDS = 1 Xw = 0.00

22

WATER XDS = 0.00 Xw = 1

DS= Dry substance (100% S , 0% NS) W = Water (Purity = 100 %) Overall M.B. : m23 = 300 ton/day = 12500 kg/h m21 = m22 + m23  m21 = m22 + 300 0.95m21 = 0 + 300  m21 = 315.78 ton/day = 13157.5 kg/h m22 = 315.78 – 300 = 15.78 ton/day = 657.5 kg/h

104

A.13 Storage balance :

(C12H22O12)

25

24

XDS = 1

STORAGE

STOREHOUSE XDS = 0.5 Xw = 0.00

Xw = 0.00 26

SPECIALTY SUGARS XDS = 0.5 Xw = 0.00

DS= Dry substance (100% S , 0% NS) W = Water (Purity = 100 %) Overall M.B. : m26 = 150 ton/day = 6250 kg/h m24 = m25 + m26  m24 = m25 + 150 0.5m24 = 150  m24 = 300 ton/day = 12500 kg/h m25 = 300 – 150 = 150 ton/day = 6250 kg/h

105

APPENDIX B : ENERGY BALANCE

B.1) Energy Balance on mixer : 2 beet o T= 25 C H1

1

mixer

(C12H22O12) o T= 65 C H2

3

(C12H22O12) o

T= 70 C H3 5

(C12H22O12) o T= 35 C H5

Water @ 25oC

H1=m1cp ∆T



Cp = Ai + BTi + CT2 ( kJ/kmol .oC) = 75.4*10-3

∆T = 25-25=0  H1=0

H2 = H2DS + H2W H2DS = m2DS CpDS ∆T CPDS at 65oC (90% purity)(15%DS) = 3.8775 kJ/kg.oC (From App D.1) H2DS = (959.58 *103)( 3.8775)(65-25) = 148830858kJ/day = 6201285.75 kJ/h   .  -3 H2W =  C dT = [75.4*10 ] [65-25] P  



= 9111034440 kJ/day = 379626435 kJ/h H2 = 9259865298 kJ/day = 385827720.8 kJ/h

106

H3 = H3DS + H3W H3DS = m3DS CpDS ∆T CPDS at 70oC (90% purity)(15%DS) = 3.88 kJ/kg.oC (From App D.1) H3DS = (1429.84*103)(3.88)(70-25) = 249650064 kJ/day = 10402086 kJ/h   .  -3 H3W =  C dT = [75.4*10 ] [70-25] P  



= 1527308055 kJ/day = 63637835.63 kJ/h H3 = 1776958119 kJ/day = 74039921.63 kJ/h H5 = H5DS + H5W H5DS = m5DS CpDS ∆T CPDS at 30oC (90% purity)(15%DS) = 3.835 kJ/kg.oC

(From App D.1)

H5DS = (1003.36*103)(3.835)(30-25) = 19239428 kJ/day = 801642.83 kJ/h   .  -3 H5W =  C dT = [75.4*10 ] [30-25] P  



= 137933618.3 kJ/day = 5747234.1 kJ/h H5 = 157173046.3 kJ/day = 6548876.93 kJ/h Q = ∑Hout - ∑Hin = (H5 + H2) – (H1+H3)

Q = 7640080225 kJ/day = 318336676.1 kJ/h

107

B.2) Energy Balance on Heater :

(C12H22O12) o T= 30 C H5

5

6

(C12H22O12) o T= 88 C H6

H5 = 157173046.3 kJ/day = 6548876.93 kJ/h H6 = H6DS + H6W H6DS = m6DS CpDS ∆T CPDS at 88oC (90% purity)(15%DS) = 3.905 kJ/kg.oC H6DS = (1003.36*103)(3.905)(88-25) = 246841610.4 kJ/day = 10285067.1 kJ/h   . [75.4*10-3] [88-25] H6W =   CPdT =







= 1500393055 kJ/day = 62516377.29 kJ/h H6 = 1747234665 kJ/day = 72801444.38 kJ/h Q = ∑Hout - ∑Hin = (H6) – (H5)

Q = 1590061619 kJ/day = 66252567.46 kJ/h

108

B.3) Energy Balance on Diffuser:

water o T=90 C H14

Beet+(C12H22O12) o T=65 C H2

2 4

Diffuser

(C12H22O12) o

T=70 C H3

(C12H22O12) o T=70 C H4

3

H2 = 9259865298 kJ/day = 385827720.8 kJ/h H3 = 1776958119 kJ/day = 74039921.63 kJ/h H14 = H14W    C dT = .  [75.4*10-3] [90-25] H14W =   P 



= 1046931278 kJ/day = 43622136.58 kJ/h H4 = H4DS + H4W H4DS = m4DS CpDS ∆T CPDS at 70oC (90% purity)(15%DS) = 3.88 kJ/kg.oC (From App D.1) H4DS = (106.6*103)(3.88)(70-25) = 18612360 kJ/day = 775515 kJ/h   . -3 H4W =  C dT = [75.4*10 ] [70-25] P  



= 113891700 kJ/day = 4745487.5 kJ/h H4=52179026.67 kJ/day = 2174126.11 kJ/h Q = ∑Hout - ∑Hin = (H3 + H4) – (H2+ H14)

Q = -8477659430 kJ/day = -353235809.6 kJ/h

109

B.4) Energy Balance on Liming : MOL(ca(OH) 2) o T=40 C H7

7

6 (C12H22O12) o T=88 C H6

8

liming

(C12H22O12) o T=88 C H8

H6 = 1747234665 kJ/day = 72801444.38 kJ/h Ca(OH)2 @ 25oC

Cp = Ai + BTi + CT2 ( kJ/kg .oC) = 89.5*10-3

  . [89.5*10-3] [40-25] H7 =   CPdT =  



= 2423746.96 kJ/day = 100989.46 kJ/h H8 = H8DS + H8W H8DS = m8DS CpDS ∆T CPDS at 88oC (90% purity)(15%DS) = 3.905kJ/kg.oC (From App D.1) H8DS = (1023.42*103)(3.905)(88-25) = 251776671.3 kJ/day=10490694.64 kJ/h H8W = 1530466938 kJ/day = 63769455.75 kJ/h H8 = 1782243609 kJ/day = 74260150.38 kJ/h Q = ∑Hout - ∑Hin = (H8) – (H6 + H7)

Q = 32585197.04 kJ/day = 1357716.54 kJ/h

110

B.5) Energy Balance on Clarifier :

(C12H22O12) T= 88oC H8

8

CLARIFIER

10

(C12H22O12) T= 88oC H10

9 SLURRY T= 88oC H9

Water @ 25oC


H8 = H8DS + H8W H8DS = m8DS CpDS ∆T CPDS at 88oC (90% purity)(15%DS) = 3.905 kJ/kg.oC (From App D.1) H8DS = (1023.42*103)(3.905)(88-25) = 251776671.3 kJ/day = 10490694.64 kJ/h

  .  [75.4*10-3] [88-25] H8W =   CPdT =  



= 1530466938 kJ/day = 63769455.75 kJ/h H8 = 1782243609 kJ/day = 74260150.39 kJ/h H9 = H9DS + H9W H9DS = m9DS CpDS ∆T CPDS at 88oC (90% purity)(70%DS) = 2.87 kJ/kg.oC (From App D.1) H9DS = (1193.997*103)(2.87)(88-25) = 215886597.6 kJ/day = 8995274.899 kJ/h

  .  -3 H9W =  C dT = [75.4*10 ] [88-25] P  



111

= 135041060.7 kJ/day = 5626710.863 kJ/h H9 = 350927658.3 kJ/day = 14621985.76 kJ/h H10 = H10DS + H10W H10DS = m10DS CpDS ∆T CPDS at 88oC (90% purity)(15%DS) = 3.905 kJ/kg.oC (From App D.1) H10DS = (767.568*103)(3.905)(88-25) = 188833241.5 kJ/day = 7868051.73 kJ/h

   C dT = . [75.4*10-3] [88-25] H10W =   P 



= 1147846773 kJ/day = 47826948.87 kJ/h H10 = 1336680015 kJ/day = 55695000.6 kJ/h Q = ∑Hout - ∑Hin = (H10 + H9) – (H8)

Q = -94635935 kJ/day = -3943163.963 kJ/h

112

B.6) Energy Balance on Heater :

10

(C12H22O12) T= 88oC H10

11

(C12H22O12) T= 92oC H11

H10 = 1336680015 kJ/day = 55695000.6 kJ/h H11 = H11DS + H11W H11DS = m11DS CpDS ∆T CPDS at 92oC (90% purity)(15%DS) (From App D.1) By Interpolation : x1 = 90 , x2 = 100 , x= 92 , y1 = 3.905 , y2 = 3.92 , y = ?

 −    =   ( −  ( ( −   



− .90  = .90  (92 − 90 (.92 (100 − 90  =3.908 CPDS at 92oC (90% purity)(15%DS) = 3.908 kJ/kg.oC H11DS = (767.568*103)(3.908)(92-25) = 200976934.8 kJ/day = 8374038.952 kJ/h

   C dT = . [75.4*10-3] [92-25] H11W =   P 



= 1220725933 kJ/day = 50863580.54 kJ/h H11 = 1421702868 kJ/day = 59237619.49 kJ/h Q = ∑Hout - ∑Hin = (H11) – (H10)

Q = 85022853 kJ/day = 3542618.875 kJ/h

113

B.7) Energy Balance on Filter :

(C12H22O12) T=92oC H11

11

FILTER

13

(C12H22O12) T=92oC H13

12 REJECT T=92oC H12

H11 = 1421702868 kJ/day = 59237619.49 kJ/h H12 = H12DS + H12W H12DS = m12DS CpDS ∆T CPDS at 92oC (90% purity)(15%DS) = 3.908 kJ/kg.oC

(By Interpolation in

previous page) H12DS = (33.368*103)(3.908)(92-25) =8736943.648 kJ/day = 364039.3187 kJ/h    C dT = . [75.4*10-3] [92-25] H12W =   P 



= 53067054.08 kJ/day = 2211127.253 kJ/h H12 = 61803997.73 kJ/day = 2575166.572 kJ/h H13 = H13DS + H13W H13DS = m13DS CpDS ∆T CPDS at 92oC (90% purity)(15%DS) = 3.908 kJ/kg.oC H13DS =(734.2125*103)(3.908)(92-25)= 192243264.2kJ/day=8010136.006 kJ/h    C dT = .  [75.4*10-3] [92-25] H13W =   P 



= 1167677963 kJ/day = 48653248.48 kJ/h H13 = 1359921227 kJ/day = 56663384.47 kJ/h Q = ∑Hout - ∑Hin = (H13 + H12) – (H11)

Q = 22356.73 kJ/day = 931.5304 kJ/h

114

B.8) Energy Balance on Evaporator : VAPOR T=100oC H14

14

(C12H22O12) T=92oC H13

13

15 Evaporator

(C12H22O12) T=95oC H15

H13 = 1359921227 kJ/day = 56663384.47 kJ/h H14 = H14W (Only vapor)    C dT = . [75.4*10-3] [100-25] H14W =   P 



= 1208245729 kJ/day = 50343572.05 kJ/h H15 = H15DS + H15W H15DS = m15DS CpDS ∆T CPDS at 95oC (90% purity)(70%DS) = 2.985 kJ/kg.oC (From App D.1) H15DS =(734.2125*103)(2.985)(95-25)=153413701.9 kJ/day=6392237.578 kJ/h   . H15W =   CPdT = [75.4*10-3] [95-25]







= 92266037.5 kJ/day = 3844418.229 kJ/h H15 = 245679745.4 kJ/day = 10236656.06 kJ/h Q = ∑Hout - ∑Hin = (H14 + H15) – (H13)

Q = 94004247.4 kJ/day = 3916843.642 kJ/h

115

B.9) Energy Balance on Crystallizer :

Water T=100oC H17

(C12H22O12) T=95oC H16

CRYSTALLIZER

(C12H22O12) T=100oC H18

H16 = H16DS + H16W H16DS = m16DS CpDS ∆T CPDS at 95oC (90% purity)(70%DS)= 2.985kJ/kg.oC (From App D.1) H16DS=(734.2125*103)(2.985)(95-25)=153413701.9kJ/day=6392237.578 kJ/h    C dT = . [75.4*10-3] [95-25] = 92266037.5 kJ/day = H16W =   P 



3844418.229 kJ/h H16T = 245679745.4 kJ/day = 10236656.06 kJ/h   . [75.4*10-3] [100-25]= 81494833.33 kJ/day H17W=   CpdT =  



= 3395618.055 kJ/h CPDS at 100oC (90% purity)(90%DS) = 2.43 kJ/kg.oC (From App A.4) H18DS = (734.2*103)(2.43)(100-25) = 133807950 kJ/day = 5575331.25 kJ/h   . H18W=   CpdT = (75.4*10-3) (100-25)= 173608500 kJ/day  



= 7233687.5 kJ/h H18T= 12809018.75 kJ/h Q = ∑Hout - ∑Hin = (H18 + H17) – (H16)

Q = 5967980.745 kJ/h

116

B.10) Energy Balance on cooler : (C12H22O12) o T=100 C H18

COOLER

(C12H22O12) o T=75 C H19

CPDS at 100oC (90% purity)(90%DS) = 2.43 kJ/kg.oC (From App D.1) H18DS = (734.2*103)(2.43)(100-25) = 133807950 kJ/day = 5575331.25 kJ/h   . (75.4*10-3) (100-25)= 173608500 kJ/day H18W=   CpdT =  



= 7233687.5 kJ/h H18T= 12809018.75 kJ/h



CpDS at 75

(90% purity)(90%DS) = 2.39 kJ/kg.oC (From App D.1)

H19DS= (734.2*103)(2.39)(75-25) = 87736900 kJ/day = 3655704.167 kJ/h   . (75.4*10-3) (75-25)= 115739000 kJ/day = H19W=   CpdT =  



= 482245.83 kJ/h H19T= 8478162.497 kJ/h Q = ∑Hout - ∑Hin = (H19) – (H18)

Q = - 4330856.253 kJ/h

117

B.11) Energy Balance on Centrifuge :

(C12H22O12) o T= 75 C H19

CENTRIFUGE

19

20

Other processes o T= 60 C H20

21

(C12H22O12) o T= 60 C H21

Water @ 25oC

Cp = Ai + BTi + CT2 ( kJ/kg .oC) = 75.4*10-3 kJ/kmol.oC

H19 = H19DS + H19W H19DS = m19DS CpDS ∆T CPDS at 75oC (92% purity)(93%DS) By Interpolation : x1 = 70 , x2 = 80 , x= 75 , y1 = 2.36 , y2 = 2.42 , y = ?

 −    =   ( −  ( ( −  − 2.6  = 2.6  (7 − 70 (2.42 (80 − 70

 =2.39

118

CPDS at 75oC (92% purity)(93%DS) = 2.39 kJ/kg.oC H19DS = (734.2*103)(2.39)(75-25) = 1754788 kJ/day = 73116.166 kJ/h    C dT = . [75.4*10-3] [75-25] H19W =   P 



= 11573900 kJ/day = 482245.833 kJ/h H19 = 13328688 kJ/day = 555362 kJ/h H20 = H20DS + H20W H20DS = m20DS CpDS ∆T CPDS at 60oC (84% purity)(94%DS) = 2.175 kJ/kg.oC (From App D.1) H20DS =(445.278*103)(2.175)(60 - 25)=33896787.75 kJ/day=1412366.156 kJ/h    C dT = . [75.4*10-3] [60-25] H20W =   P 



= 4166.981 kJ/day = 173.624 kJ/h H20 = 33900954.73 kJ/day = 1412539.78 kJ/h H21 = H21DS + H21W H21DS = m21DS CpDS ∆T C P=4.187− DS (0.0297−4.6.10−5 P )+7.5*10−5 DS T (1.18)

by integral the equation from (25 - 60

o

C) we find CPDS∆T(100%

purity)(95%DS) = 68.24 kJ/kg H21DS = (300*103)(68.24) = 23415000 kJ/day = 975625 kJ/h   . [75.4*10-3] [60-25] H21W =   CPdT =







= 2313523.33 kJ/day = 96396.8055 kJ/h H21 = 22784501.46 kJ/day = 949354.23 kJ/h Q = ∑Hout - ∑Hin = (H20 + H21) – (H19)

Q = 43356768.19 kJ/day = 1806532.008 kJ/h

119

B.12) Energy Balance on Dryer :

(C12H22O12) o T= 60 C H21

DRYER

21

23

(C12H22O12) o T= 65 C H23

22

WATER o T= 65 C H22

Water @ 25oC


H21 = H21DS + H21W H21DS = m21DS CpDS ∆T CPDS at 60oC (100% purity)(95%DS) We can calculate the specific heat of pure sucrose solutions by using the following equation (Bubnik et al. 1995): 5 C P= 4.187−DS (0.0297−4.6.10 P )+7.5*10 −

−5

DS T (1.18)

DS Dry substance content (for pure sucrose solutions, DS = S) T Temperature (oC) P Purity (for pure solutions, P = 100)

by integral the equation from (25 - 60

o

C) we find CPDS∆T(100%

purity)(95%DS) = 68.24 kJ/kg H21DS = (300*103)(68.24) = 20470978.13kJ/day = 852957.42 kJ/h   . [75.4*10-3] [60-25] H21W =   CPdT =







= 2313523.33 kJ/day = 96396.8055 kJ/h H21 = 22784501.46 kJ/day = 949354.23 kJ/h

120

H22 = H22W CPW at 65oC (0%DS) = 3.016 kJ/mol .oC    C dT = . [75.4*10-3] [65-25] H22W =   P 



= 2644026.667 kJ/day = 110167.778 kJ/h H22 = H22W= 2644026.667 kJ/day = 110167.778 kJ/h H23 = H23DS H23DS = m23DS CpDS ∆T CPDS∆T from (25 – 65oC) (100% purity)(100%DS) = 74.16 kJ/kg

(From

equation) H23DS = H23 = (300*103)(74.16) = 22248000 kJ/day = 927000 kJ/h Q = ∑Hout - ∑Hin = (H22 + H23) – (H21)

Q = 2107525.207 kJ/day = 87813.55 kJ/h

121

B.13) Energy Balance on Cooler :

(C12H22O12) o T= 65 C H23

23

COOLER

24

(C12H22O12) o T= 30 C H24

H23 = H23DS H23DS = m23DS CpDS ∆T CPDS∆T from (25 – 65) (100% purity)(100%DS) = 74.16 kJ/kg

(From

equation) H23DS = (300*103)(74.16) = 22248000 kJ/day = 927000 kJ/h H24 = H24DS H24DS = m24DS CpDS ∆T CPDS∆T from (25 – 30) (100% purity)(100%DS) = 8.5 kJ/kg (From equation) H24DS = (300*103)(8.5) = 2548687.5 kJ/day = 106195.31 kJ/h Q = ∑Hout - ∑Hin = (H24 – H23)

Q = -19699312.5 kJ/day = - 820804.69 kJ/h

122

APPENDIX C : Mass balance of water recycling and treatment

C.1 Mass balance of washing water recycling : m1= 3950 ton/day (beet) m2= 7900 ton/day (water) m2(recycled water) = 5530 ton/day m2 (fresh water) = 7900-5530= 2370 ton/day m3= 3554 ton/day (beet after washing) m4= 8296 ton/day (waste water) m4(mud)= m1- m3 = 3950-3554= 396 ton/day m4(water) = m4- m4(mud)= 8296-396= 7900 ton/day m5= 0.67(8296) = 5530 ton/day (recycled water) m6= 0.33(8296)= 2766 ton/day (water +mud) m6(water)=m6-m(mud)= 2766-396= 2370 ton/day m7= 0.7(2370)= 1659 ton/day (water) m8= 0.3(2370)+396= 1107 ton/day (water +mud)

C.2 Mass balance of Diffusion Tower water recycling :

m9= 710.8 ton/day ( pulp) m9(water)= 0.9(710.8)= 639.72 ton/day m9(pulp)= 0.1(710.8)= 71.08 ton/day m10= 0.8(639.72)= 511.78 ton/day (water recycled) m11= m9- m10= 710.8-511.78=199.02 ton/day m11(water)= 0.2(639.72)= 127.94 ton/day m11(pulp)= m9(pulp)=m12(pulp)= 71.08 ton/day

123

APPENDIX D (Physical properties of sugar ) D.1) Heat capacity :

124

125

SUGAR PRODUCTION.pdf

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