Yogurt Production Our Thesis Complete -1

YOGURT PRODUCTION FROM LACTIC ACID FROM BACTERIA

NURFATIN AMIRAH BINTI IZHAB (2012888124) MOHD NAZMIE BIN MOHAMED MOKHTAR (20128801260 HAZIRAH BINTI HAFIZ (2012434468) MUSALMAH BINTI ADANAN (2012218062)

FACULTY OF CHEMICAL ENGINEERING UNIVERSITI TEKNOLOGI MARA SHAH ALAM JUNE 2013 1

DECLARATION

“I hereby declare that this report is the result of my own work except for quotes and summaries which have been duly acknowledged.”

---------------------------------------NAME: NURFATIN AMIRAH BINTI IZHAB ID

DATE: 10/6/2013

: 2012888124

---------------------------------------NAME: MOHD NAZMIE BIN MOHAMED MOKHATR ID

DATE: 10/6/2013

: 2012880126

---------------------------------------NAME: MUSALMAH BINTI ADANAN ID

DATE: 10/6/2013

: 2012218062

---------------------------------------NAME: HAZIRAH BINTI HAFIZ ID

DATE: 10/6/2013

: 2012434468

2

SUPERVISOR’S CERTIFICATION

“I hereby declare that I have read this thesis and in my opinion this project report is sufficient in terms of scope and quality for the award of Bachelor in Chemical Engineering (Hons).”

Signature

:______________________

Name

: Nur Shahidah Binti Ab. Aziz

Date

:______________________

3

Accepted:

Signature : _______________ Date

:________________ Head of programme Dr. Jefri Faculty of Chemical Engineering Universiti Teknologi MARA Shah Alam

Signature :________________ Date

:________________ Coordinator Miss Nur Shahidah bt Abd Aziz Faculty of Chemical Engineering Universiti Teknologi MARA Shah Alam

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Table of Contents LIST OF TABLES .............................................................................................................................................. 6 LIST OF FIGURES ............................................................................................................................................ 6 LIST OF EQUATIONS ...................................................................................................................................... 7 LIST OF ABBREVIATION ................................................................................................................................. 8 LIST OF SYMBOLS .......................................................................................................................................... 8 CHAPTER ONE: GENERAL REVIEW ................................................................................................................ 9 1.1

Introduction ...................................................................................................................................... 9

1.2

Process Involved ............................................................................................................................. 11

1.2.1

Process Flowchart ....................................................................................................................... 11

1.2.2

Process and Reaction Description............................................................................................... 12

1.3

Thermodynamics Properties of Raw Materials and Products ........................................................ 15

1.4

Waste generation and Environmental Act ...................................................................................... 17

1.5

Conclusion ....................................................................................................................................... 18

CHAPTER TWO: PROCESS FLOW AND DESCRIPTION .................................................................................. 19 2.1

Process Assumptions ...................................................................................................................... 19

2.2

Process Flow Diagram ..................................................................................................................... 21

2.3

Stream Tables.................................................................................................................................. 22

2.4

Equipments Tables and Description ............................................................................................... 23

2.4.1

Quantity, Quality Control and Storage ....................................................................................... 25

2.4.2

Materials and Energy Balance..................................................................................................... 26

2.4.3

Heat Exchanger ........................................................................................................................... 39

2.4.3.1

Heat Transfer Mode, Type flow and Calculations ....................................................................... 39

2.5

Bioprocess and Metabolic Regulations ........................................................................................... 53

2.5.1

Biomolecules Involved ................................................................................................................ 53

2.5.1.1

Lactose ........................................................................................................................................ 53

2.5.1.2

Glucose........................................................................................................................................ 54

2.5.1.3

Galactose..................................................................................................................................... 55

2.5.1.4

Lactase ........................................................................................................................................ 57

2.5.2

Biochemical Pathway .................................................................................................................. 57

CHAPTER THREE: CONCLUSION AND RECOMMENDATIONS ...................................................................... 65 REFERENCE .................................................................................................................................................. 66 5

LIST OF TABLES Table 1: Stream table for continuous process of yogurt, streams 1-20 ....................................................... 22 Table 2: Equipment table for volumetric flow meters ................................................................................ 23 Table 3: Equipment table for temporary storage tank................................................................................. 23 Table 4: Equipment table for fermenter ...................................................................................................... 23 Table 5: Equipment table for filter .............................................................................................................. 24 Table 6: Equipment table for centrifuger .................................................................................................... 24 Table 7: Equipments table for pump .............................................................. Error! Bookmark not defined. Table 8: Equipment table for mixers........................................................................................................... 24 Table 9: Equipment table for homogenizer ................................................................................................ 25 Table 10: Equipment table for heat exchangers ............................................. Error! Bookmark not defined. Table 11: Equipment table for storage freezer ............................................................................................ 25 Table 12: Heat transfer properties at heat exchanger .................................................................................. 39

LIST OF FIGURES Figure 1 : Flowchart showing proposed process for yogurt production from lactic acid. .......................... 11 Figure 2: Hydrolysis of Sucrose (Averill & Eldredge, 2013) ..................................................................... 16 Figure 3: Lactic acid fermentation (Farabee, 2010) .................................................................................... 16 Figure 4: Filtration mass balance ................................................................... Error! Bookmark not defined. Figure 5 Centrifuge Mass Balance ................................................................. Error! Bookmark not defined. Figure 6 Centrifuge Energy Balance .............................................................. Error! Bookmark not defined. Figure 7 Mixer M-101 mass balance ............................................................. Error! Bookmark not defined. Figure 8 Mixer M-102 mass balance ............................................................. Error! Bookmark not defined. Figure 9 Homogenizer mass balance ............................................................. Error! Bookmark not defined. Figure 10 Homogenizer mass balance ........................................................... Error! Bookmark not defined. Figure 11: Temperature distribution of a counter flow of heat exchanger ..... Error! Bookmark not defined. Figure 12 Pasteurizer mass balance ............................................................... Error! Bookmark not defined. Figure 13 Pateurizer energy balance .............................................................. Error! Bookmark not defined. Figure 14 Fermenter mass balance................................................................. Error! Bookmark not defined. Figure 15 Mass balance at storage tank ......................................................... Error! Bookmark not defined. Figure 16: Chemical structure of lactose (Calvero, 2013) .......................................................................... 53 Figure 17: Chemical structure of glucose (Nave, 2012) ............................................................................. 54 Figure 18: Hemiacetal functional group in glucose (Monosaccharide-Structure of Glucose, 2001) .......... 55 Figure 19: Molecular structure of galactose (Ophardt, Galactose, 2003) ................................................... 56 6

Figure 20: Difference between galactose and glucose in structure (Ophardt, Galactose, 2003) ................ 56 Figure 21: Conversion of lactose to galactose and glucose (Taylor & Stahlberg, 2005) ............................ 57 Figure 22: Overview of glycolysis (Glycolysis, 2013) ............................................................................... 58 Figure 23: Phosphorylation of glucose (Helmenstine, 2013)...................................................................... 59 Figure 24: Conversion of glucose-6-phosphate to fructose-6-phosphate (Helmenstine, 2013) .................. 59 Figure 25: Phosphorylation of fructose-6-phosphate (Helmenstine, 2013) ................................................ 60 Figure 26: Cleavage of fructose-1,6-phosphate (Helmenstine, 2013) ........................................................ 60 Figure 27: Interconversion of glyceraldehaydes-3-phosphate and dihydroxyacteone phosphate ............... 61 Figure 28: Oxidation of glyceraldehyde-3-phosphate................................................................................. 61 Figure 29: Phosphoryl group transfer ......................................................................................................... 61 Figure 30: Interconversion of 3-phosphoglycerate to 2-phosphoglycerate................................................. 62 Figure 31: Dehydration of phosphoenolpyruvate ....................................................................................... 62 Figure 32: Synthesis of pyruvate ................................................................................................................ 62 Figure 33: Galactose metabolism................................................................................................................ 63 Figure 34: Lactic acid fermentation ............................................................................................................ 64

LIST OF EQUATIONS Equation 1: Chemical equation of glucose to pyruvate (Ophardt, Glycolysis Summary, 2003) ................ 16 Equation 2: Chemical equation of pyruvate to lactate (Robergs, 2001) ..................................................... 17 Equation 3: Overall reaction of glycolysis (Ophardt, Glycolysis Summary, 2003) ................................... 58

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LIST OF ABBREVIATION Abbreviation LAB Sp. OHTC Re Nu NADH DHAP NAD

Lactic Acid Bacteria Species Overall Heat Transfer Coefficient Reynolds Nusselt Reduce nicotinamide adenine dinucleotide Dihydroxyacetone phosphate Nicotonamide adenine dinucleotide

LIST OF SYMBOLS Symbol °C α β µ Δ ∑ Cp ΔTlm h Q W U v P T

Degree Celcius Alpha Beta Viscosity Changes Summation Specific Heat capacity at constant Pressure Temperature log mean Enthalphy Heat transfer Work Internal energy Specific volume Pressure Temperature

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CHAPTER ONE: GENERAL REVIEW 1.1

Introduction

Yogurt is known longer than we can imagine which is since 6000 B.C. Even the Mongol Empire lead by Genghis Khan lived on yogurt. However, the first industrialized yogurt is in the year of 1919 in Barcelona by Isaac Carasso before the goodness concealed in yogurt being known generally to public. Nowadays, people have started to realize the important of yogurt in their everyday life. Yogurt gives a lot of nutrition to our body and also helps the circulation process in our body to run well. It is an alternative or another milk substitutes for those who are lactose intolerant. Due to this growing awareness, their demand towards yoghurt production has automatically increases. The suitable storage temperature for yoghurt is 7.2◦C and below. This is due to the presence of living microorganism in the yogurt which is the lactic acid bacteria where the temperature is set to inhibit them from undergo fermentation that might cause the yogurt become more acidic. The lactic acid bacteria that usually used in the industries for yogurt production are Lactobacillus bulgaricus, Lactobacillus delbruecki sp. and Streptococcus thermophillus each with optimum temperature of 45◦C (Todar). The composition of the yogurt is also different depending on the type of yogurt. For regular yogurt, the fat and milk solid content are at least higher than 3.25% and 8.25% respectively whereas for low-fat yogurt, the fat content is in between 0.5% and 2%. There is also non-fat yogurt which composes of less than 0.5% of fat. Both of the low-fat and non-fat yogurt have the same milk solids composition with the regular yogurt. (Milk Processing-Yoghurt Production, 2013). Particularly, solid content of milk up to 16% of total mass, 1-5% of fat and 11-14% of solid non-fat (SNF) (Watson, 2013) The pH of the yogurt usually maintained at pH 3 or pH 4 which occur during the fermentation where the lactic acid bacteria lower the pH from 6.5-6.6 to the desired pH. The yogurt must be at least at pH of 4.4 to be legally sold in the United States. (Choosing a Yogurt Starter Culture)

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The processes that take place for yogurt production varies depending on the types of yogurt. Yogurt actually comes in wide variety as the flavors, forms and textures are also varies. However, generally, there are three types of yogurt which are low-fat, non-fat and regular yogurt which each of them varies in their composition. Thus, the processes are slightly different to ensure their composition is well fixed. The process also depends on the style as it varies on how they are made. The three main style of making are Balkan-style, Swiss style and Greek style yogurt. The Balkan-style or common known as set-style yogurt usually used to produce plain yogurt. It has thick texture and suitable usage for recipes. The Swiss style has slightly lighter texture with the adding of flavors and fruits. It commonly use in the industry nowadays. The Greek style has a very thick textures and is made by either evaporate water from the milk or straining whey from a plain yogurt to produce creamier taste. It tends to hold up during heating, thus make it suitable for cooking too. By considering all the major existing process, new process flow is suggested in this project for the production of yogurt from lactic acid bacteria.

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1.2

Process Involved 1.2.1 Process Flowchart Filtration Centrifugation Mixing Powder Skimmed milk Stabilizer

Heat Treatment Homogenization Pasteurization Cooling Fermentation Cooling Mixing

Stabilizers and Flavoring

Cooling Packaging and Storage

Figure 1 : Flowchart showing proposed process for yogurt production from lactic acid.

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1.2.2 Process and Reaction Description Figure 1 is the flowchart that shows proposed process to apply in the production of yogurt after a few existing processes were revised. Each of the process functions and how they will affect the end-product are also considered in the process flow suggestion. Basically, raw milk usually being filtered first to prevent any impurity in the milk that can cause any harm to the yogurt production or the consumer. Some of the factory existed, preheated the milk to kill any microorganism present in the milk to avoid any unneeded reaction. However, despite heating it first, centrifugation is done first in the process flow suggested to minimize the energy usage as before fermentation is done, a pasteurization process will be needed to completely kill the other microorganism. The centrifugation and homogenization process are the combo for the standardization and modification of the milk. These steps are essential to produce a good quality end-product but more importantly, the steps will provide the best condition for fermentation to occur later. The other existing process included evaporation as one of the process to standardize the milk. The reason is to increase the mass percentage of milk in the mixture or in other word, to remove the water. Unfortunately, evaporater do consumed a lot of energy, thus in the suggested flow process, evaporation process is replaced by adding powder skimmed milk to increase the mass percentage. Centrifugation process is usually used in the industries to separate fat from the milk in order to lower the fat content in the product. The type of centrifuge used for milk usually disc-bowl centrifuge. The revolutions per minute (rpm) of the centrifuge ranging in between 2000- 7000 rpm for fat to separated from the milk (HYFOMA). The centrifuged milk was then mixed with powder skimmed milk and stabilizers to increase the mass percentage and maintain the mixture from coagulate. The mixture undergo homogenization after their temperature is increased by heat treatment ranging from 55-75◦C. The heat treatment is needed to favor the process of homogenization to occur.

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Homogenization process which is the last step before the milk is ready to ferment, is needed to form better texture and releasing composition that will stimulate the starter culture. It is a process where the fat globules are being broken down by forcing the milk to go through small opening under high pressure. The pressure usually varies in between 100-200 atm for milk homogenization in yogurt production. After the homogenization is done, sample of the milk is taken to ensure that the composition is suitable for the next process. Next is fermentation of the milk, but the readily milk must undergo pasteurization first to kill the microorganism in it and only then the temperature is lowered to provide the best condition for fermentation. The pasteurization is done under high temperature for a short time, only enough to kill the microorganism. For some other existing process, they usually pasteurize first before the homogenization to not only kill the bacteria but also to denature the whey protein. However, in this process, the pasteurization is needed only to kill the microorganism. After the pasteurization is done, the milk must be cool down to 42-46◦C and the same temperature is maintained during the fermentation as it is the most optimum range of temperature for the selected lactic acid bacteria. The lactic acid bacteria also plays significant role in the yogurt production so that the fermentation will develop without bringing any harm to the product as well as the consumer later. The duration of the fermentation is regularly 3-4 hours. By that time, the pH of the milk initially at 5.0 to 6.6 will dropped to at least pH 4.0 by the presence of lactic acid converted by the LAB. As the pH lowered down, the protein inside the milk will denatured and stick together forming the better texture of yogurt. (Yoghurt Production, 2013) To stop the activity of the live culture after the fermentation, the product which is the raw yogurt will be cooled down to at least 5- 7◦C. This is crucial as further fermentation will give the yogurt extra sour taste due to excessive accumulation of the lactic acid. If this occurs, the yogurt taste is spoiled and might be off from marketed. The raw yogurt is then, will be mixed together with stabilizers and flavor before the end product is ready for packaging. The flavor and the fruits are needed to enhance the taste while the stabilizers are added to maintain the firmness, jelly-like form and increasing the texture quality of the yogurt. Common stabilizers are gelatin, pectin, agar and starch. (Watson, 2013) Here, there are two types of way where the flavors and fruits can be added. First by using the set-style 13

by adding the fruit at the bottom of the cup and the inoculated yogurt are poured later during the packaging or using the Swiss-style or stirred-style to blended the fruit together with the cooled yogurt prior to packaging. (Milk Processing-Yoghurt Production, 2013) Swiss style is found to be more suited for industries so that the yogurt is well mixed together with the stabilizers. For the packaging, there are high possibilities for contamination to happen without proper prevention. The usual type of contamination to happen is cross contamination but it is preventable. Some of the methods of prevention are such as keeping the plant design and production flow minimize from any likelihood of cross-contamination (ex: employees working in raw processing area should not access RTE area), clean filtered air, cleaning and sanitation of equipments regularly, separate the storage of raw materials and product and others. (Cross Contamination, 2003)

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1.3

Thermodynamics Properties of Raw Materials and Products

In the production of yoghurt from bacteria, bacteria used in this production of yoghurt are Lactobacillus Bulgaricus and Streptococcus thermophillus.

These bacteria undergo two

biochemical processes which are hydrolysis and fermentation in order to produce lactic acid. The first reaction occurs when sucrose is converted to glucose and fructose. This process is known as hydrolysis process which is catalyzed by enzyme sucrase provided by the bacteria (H.Garret & Grisham, 2010). The temperature of the culture tank is between 70°C to 80°C as it is the optimum temperature for the enzyme to react (Heinen, 1970). The optimum pressure of the tank is 1 atm. The sucrose and the enzyme appear as liquid in this tank. Sucrose’s heat capacity is calculated using Kopp’s rule, a simple empirical method for estimating the heat capacities. (Cp)C12H22O11 = 12(Cpa)C + 22(Cpa)H + 11(Cpa)O = 12(12) + 22(18) + 11(25) = 815 J/mol °C = 0.815 kJ/mol °C Sucrose has a density of 1.59g/cm3 (Density of Sucrose, 2013). The melting point of sucrose is 367°F. Sucrose does not have a boiling point as I break down to form caramel before boils (Boiling Point of Sucrose, 2013). The culture tank is an open system tank where there are changes of heat and matter that occurs. This is a steady state flow system. The heat is absorbed in this reaction in order to break the bond of sucrose to produce glucose and fructose. Thus, q > 0 as heat energy is needed in the bond breaking.

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Figure 2: Hydrolysis of Sucrose (Averill & Eldredge, 2013)

The second process is lactic acid fermentation. Glucose is converted to lactate in this process. The product of this reaction is lactic acid and NAD.

Figure 3: Lactic acid fermentation (Farabee, 2010)

There are two main phases in lactic acid fermentation which are the conversion of glucose to pyruvate and the conversion of pyruvate to lactic acid. C6H12O6 + 2 NAD+ + 2 ADP + 2 P -----> 2 pyruvic acid, (CH3(C=O) COOH + 2 ATP + 2NADH + 2 H+ Equation 1: Chemical equation of glucose to pyruvate (Ophardt, Glycolysis Summary, 2003)

Pyruvic acid + NADH + H+

lactic acid + NAD+

lactate-Na+ + NAD+ + H+

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Equation 2: Chemical equation of pyruvate to lactate (Robergs, 2001)

The fermentation tank’s temperature is kept between 42-46°C as these range of temperature are optimum for the bacteria used. The pressure of the tank is kept constant at 1 atm. This glucose is in liquid phase. In a fermentation tank milk is ferment with the bacteria as one of the procedure to produce yoghurt. The milk which enters the fermentation tank has a specific heat capacity of 3.22 kJ kg-1 °C-1. The boiling point of the milk is around 100°C as milk is mostly water (Tamara, 2007). For the melting point of the milk, it is above -0.250°C (Tamara, Freezing Point of Milk, 2007). Skimmed milk is said to have the density of 1.026 kg/L at 38.9°C. The density changes as the lighter the milk fat rises to the surface (Elert, 2002). Glucose has a density of 1.54g/cm3 (Glucose, 2013). The heat specific heat capacity is 155J/K (Schroeder, V, & Wesley, 2000). The usual boiling point of glucose is around 150°C and the melting point is 146°C (Boiling Point of Glucose, 2013). Impurities lower the glucose’s melting point (Melting Point of Glucose, 2013). This reaction is a steady state flow and an open system reaction as there is a change in form of heat and matter. As NAD is also the product in lactic acid fermentation, the reaction is an exothermic process. Energy is released in this reaction in form of heat, q < 0.

1.4

Waste generation and Environmental Act

In this yogurt production, waste product is being disposed from the system during the filtration process. The idea of this process is to increase the creaminess of the frozen yogurt, the amount of protein and calcium in the product and to decrease the amount of lactose. To achieve this, a volume reduction factor of 4.55 is needed (Premaratne and Cousin, 1991; pg. B-2). To do so, only 78% of the incoming skim milk is filtered and only 22% of the skim milk becomes UF milk. The cold filtrate can be used to cool the compressed ammonia, grow the bulk culture and even sold as pig feed (Knight,2008). The working fluid used in this production is water. The 17

water is reused for the same purpose as water is renewable. Also, water is easily found and cheap. This can reduce the cost of the production. There are three major safety hazards associated with frozen yogurt manufacturing; microbiological, chemical and physical. The greatest hazard is microbiological, which may affect the human health. If the design parameters are not strictly controlled, potential risks may occur throughout the process from milk receiving to storage and transportation. Chemical hazards are a concern as we are dealing with large quantities of toxic, highly corrosive compound onsite. Physical hazard can result in human injury, or worse, fatality. This hazard inflicts direct impact on the personnel working at the facility during the operational phase.

1.5

Conclusion

Yogurt production varies in the process of making as well as the textures of the end-product. Process flow, the equipments and the culture and raw materials must be chosen depending on the need for the type of the yogurt end-product. Process flow must be suitable so that the raw materials don’t lose its texture, viscosity and the nutrient itself. This is because some of the existing process can affect the materials and chemical composition. The way of handling the equipments involved in the process especially at crucial tank such as fermenter can make a big loss if there is no turning back or restoration if there are any mistakes happen. For example, the pH exceeded the desired pH due to lactic acid production form way too many. Besides that, the types of culture, as well as the raw materials also need to be chosen precisely for the reaction to happen accordingly.

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CHAPTER TWO: PROCESS FLOW AND DESCRIPTION 2.1

Process Assumptions

The process to make yogurt is described in this section. A block flow diagram of the process can be found in section 2.2, Figure 2 and the process flow diagram can be found in section 2.3, Figure 3. The stream tables are given by Table 1. The equipment tables are located in section 2.5. Later in the same section, detailed mass and energy balance as well as the calculations for this process can be found. From the process, a few assumptions are needed to simplify the calculation and estimation of the product mass and energy balance as well as the heat transfer calculation. The assumptions are: 1. The yogurt production process is steady-flow at each component. 2. During the heat exchange at each tank and stream, the heat loss to surrounding is considered negligible. 3. All the process systems are assumed to open-system. 4. The kinetic and potential energy, KE and PE are assumed negligible. 5. To avoid any corrosion, or other impurities from contaminate during the process, all the equipments is assumed are made of stainless steel materials. 6. The water and steam stream is assumed not to leak. 7. The basis for the whole production is assumed 3000 kg of raw milk is being processed per day. 8. The pressure at each tank except the homogenizer is assumed to be at 1 atm. There are also specific assumptions at selected stream and equipments which based on process flow diagram in section 2.2.

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Stream

5 7

10

14

Assumptions

1. 2. 1. 2. 3. 1. 2. 3. 1. 2. 3.

15-19

4. 1. 2. 3.

Equipments FL-101 CF-101

HG-101

F-101

E-101 E-102 E-103 E-104

The flow of liquid is steady-flow The filter completely filtered impurities Steady-flow process Whey protein and undesired fat composition are completely removed after centrifugation. The pressure is assumed 1 atm The homogenizer is assumed single-phase homogenizer. Steady-flow process The mass is conserved in homogenizer. The fermenter is assumed as open system. The energy is conserved in the fermenter due to constant temperature. The composition is assumed conserved even though the textures become more jelly-like. Steady-state during the fermentation process Steady-flow process. The properties of milk and yogurt entering the heat exchanger are considered the same as water. Heat loss to surrounding is considered negligible.

Assumption

1. 2. 1. 2. 3. 1. 2. 3. 1. 2. 3. 4. 1. 2.

Steady-flow process Impurities are completely removed. Steady-flow process Heat loss to surrounding is negligible Undesired composition is assumed removed. Steady-flow process Mass is assumed conserved No heat loss to surrounding where it’s negligible. Steady-flow process Assumed as open system. Energy is assumed conserved. Mass is assumed conserved. Steady flow process Milk and yogurt properties are assumed have the same properties with water. 3. Mass is conserved, no composition change. 4. Average constant thermal properties (thermal conductivity and specific heat) and convective heat transfer coefficient along the heat exchanger. 5. Negligible internal heat generation and negligible free convection 6. Average temperature is taken for measurement.

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2.2

Process Flow Diagram

Figure 4: Process flow diagram

Streams number 21

FM-101 ST-101 ST-102 FL-101 CF-101 E-101 E-102 E-103 E-104 HG-101 M-101 M-102 CT-101 F-101 P-101

Volumetric Flow meter Temporary storage tank Filter Centrifuge Heat Exchanger

Homogenizer Mixer Culture Tank Fermenter Pump

2.3

Stream Tables

Table 1: Stream table for continuous process of yogurt, streams 1-21

Stream Temperature (◦C) Pressure (atm) Mass flow (kg/day) Component Stream Temperature (◦C) Pressure (atm) Mass flow (kg/day) Component

1 4 1 3000 Raw milk 6 50 1 66.07 Undesired Composition

Stream 13 Temperature (◦C) 45 Pressure (atm) 1 Mass flow (kg/day) 105.87 Component Culture inoculated with NFDM

Stream Temperature (◦C) Pressure (atm) Mass flow (kg/day) Component

18-20 35.2 1 122 Working fluid

2 4 1 3000 Raw milk 7 65 1 2930.93 Milk

14 30 1 3302.43 Raw Yogurt

21 27 1 120.07 Proline Aspartame

3 4 5 4 27 4 1 1 1 3000 3 2997 Raw milk Impurities Milk 8 9 10-12 92 45 40 1 1 1 265.63 3196.56 3196.56 Proline Concentrated Concentrated skimmed Milk milk milk (HE stream) 15 16 17 30 5 32.5 1 1 1 3302.43 3422.5 3422.5 Raw Yogurt Cooling and Yogurt storage (HE stream) 22&23 65 1 121.7749 Working fluid

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2.4

Equipments Tables and Description

Table 2: Equipment table for volumetric flow meters

Volumetric flow meter FM-101 MOC* SS Type Magnetic Inductive Component Milk Inlet Temperature (◦C) 4 Inlet Pressure (atm) 1 Mass flow (kg/day) 3000

Table 3: Equipment table for temporary storage tank

Storage tank MOC* Type Component Inlet Temperature (◦C) Inlet Pressure (atm) Mass capacity (kg/day)

ST-101 SS Cone roof Raw milk 4 1 3000

ST-102 SS Cone roof Raw milk 4 1 3000

Table 4: Equipment table for fermenter

Fermenter MOC* Type Component Temperature (◦C) Pressure (atm) Volume (m3) Mass capacity (kg/day) Component

F-101 SS Plug flow Milk mixture and Bulk Culture 45 1 4 3500 Raw milk

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Table 5: Equipment table for filter

Filter MOC* Type Component Inlet Temperature (◦C) Inlet Pressure (atm) Outlet Pressure (atm) Mass flow in (kg/day) Mass flow out (kg/day) Filtrate flux (kg/day) Area (m2)

FL-101 SS Nylon-filter Raw milk 4 1 1 3000 2997 3 27.63

Table 6: Equipment table for centrifuge

Centrifuger MOC* Type Mass capacity (kg/day) Component Temperature (◦C) Pressure (atm) Revolution per minute (rpm)

CF-101 SS Disc bowl centrufger 3000 Raw milk 50 1 7000

Table 7: Equipment table for mixers

Mixers MOC* Type Component

Inlet Temperature (◦C) Inlet Pressure (atm) Mass capacity (kg/day) Mixing time (hr) Volume (ft3)

M-101 SS Closed vessel with agitator Raw milk Powder skimmed milk Stabilizer (Proline) 50 1 3500 0.5 2

M-102 SS Closed vessel with agitator Raw yogurt Stabilizer (Proline) Aspartame 30 1 3500 0.5 2

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Table 8: Equipment table for homogenizer

Homogenizer MOC* Type Component Temperature in (◦C) Temperature out (◦C) Inlet pressure (atm) Pressure (atm)

FM-101 SS Single stage Mixture of milk 50 65 1 178

Table 9: Equipment table for storage freezer

Storage freezer MOC* Component Inlet temperature (◦C) Outlet temperature (◦C) Pressure (atm) Mass flow (kmol/hr) Heat duty (kW)

SF-101 SS Yogurt 64 37 1 121.77 4.56

2.4.1 Quantity, Quality Control and Storage When the raw milk arrives at the plant, the quantity of milk delivered is determined by sending the milk through volumetric flow meter, FM-101, on its way to temporary storage tank. The mass of the milk delivered is determined from the density of the milk through the volumetric flow meter reading. Before any other, filtration was done to remove impurities which in this case, we use nylon-filtered tank. Only then, the milk is sent for the real production of yogurt processes. The temporary storage tanks are needed as not all of the raw milk will be used once they arrived at the plant.

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2.4.2 Materials and Energy Balance In yogurt production, there are five main stages not including heat treatment. They are filtration, centrifugation, mixing, homogenization and fermentation. In the production, 3000kg/day of raw milk processes is used as basis. Filtration is to remove all the impurities such as dust and hair to avoid contamination to final product. It is assumed that the composition of impurities in raw milk is 0.1% and during filtration, all of them are removed. FL-101 M2 = ________ kg/day Ximpurities = 1

M3 = _________kg/day Xmilk = 1 Ximpurities = 0

M1 = 3000 kg/day Xmilk = 0.999 Ximpurities = 0.001 FILTER

M1 = M2 + M3 3000 = M2 + M3

(kg/day) (kg/day)

Milk mass fraction: (0.999)(3000) = (0)M2 + (1)M3 M3 = 2997

(kg/day) (kg/day)

Impurities mass fraction: (0.001)(3000)= (1)M2 + (0)M3 M2 = 3

(kg/day) (kg/day 26

After the filtration, the milk is sent to centrifuge to remove undesired fat content and whey protein. Below is the table of raw cow milk composition.

Composition of milk Water Lactose Fat Whey protein Protein Other

% 86.5 4.8 4.5 0.9 2.6 0.7

Table 10: Raw milk composition

The desired milk composition in this production that we want to achieve is 0% whey protein and 0.0325% of fat from total mass fraction of the milk. In below block diagram, lactose, protein and other are assumed to be solid composition. At the filtrate out stream, by using ratio, mass fraction of filtrate removed is composed of 0.58 of fat and 0.42 of whey protein. CF-101

M4 = ________ kg/day Xfat_2 = 0.58 Xwhey_2 = 0.42 M5 = _________ kg/day Xwater_5 = (x) Xsolid_5 = (1-0.0325-x) Xfat_5 = 0.0325 Xwhey_5 = 0

M3 = 2997 kg/day XSolid_3 = 0.081 Xwhey_3 = 0.009 XWater_3 = 0.865 Xfat_3 = 0.045

CENTRIFUGER

M3 = M4 + M5 2997 = M4 + M5

(kg/day) (kg/day)

27

Water mass fraction: (0.865)(2997) = (0)M4 + (x)M5 (x)M5 = 2592.41

(kg/day) (kg/day)

Solid mass fraction: (0.081)(2997) = (0)M4 + (0.9675-x)M5 (0.9675-x)M5 = 242.76

(kg/day) (kg/day)

By comparing equation from water and solid mass fraction balance: x= 0.8845 Thus, mass fraction of water = 0.8845 mass fraction of solid = 0.083 M5 = 2930.93 M4 = 66.07

(kg/day) (kg/day)

After the whey and undesired fat remove, the milk solid content need to be increase at least to 16% of total mass of the milk. Thus, considering fat is also included in the solid composition, the total solid mass composition entering the mixer is 11.35%. Therefore, at least another 4.65% of solid mass of milk is needed to produce optimum solid composition. There are two ways which are evaporating the water or adding skimmed powder milk. In this case, we use skimmed powder milk. To do so, proline is also added as the stabilizers. For the first stage mixing, only 0.5% mass fraction from total mass of milk of proline is needed to stabilize the milk. The proline will also be considered to be included in solid composition. We assumed that the outlet will atleast compose of 4.15% of skimmed milk powder and 0.005% of proline from total mass mixed milk. Total mass fraction of proline and skimmed milk powder is also calculated by ratio of mass composition needed to increase the total solid mass in milk.

28

M-101

M6 =___________ kg/day Xproline = 0.11 Xs.milk = 0.89 M7 = __________kg/day Xwater = x Xs.milk = 0.0415 Xproline = 0.005 Xsolid = (1-x-0.04150.005)

M5 = 2930.93 kg/day Xwater = 0.8845 Xsolid = 0.1155

MIXER

M5 + M6 = M7 2930.93 + M6 = M7

(kg/day) (kg/day)

Mass fraction of water: (0.8845)(2930.93) + (0)M6 = (x)M7 xM7 = 2592.41

(kg/day) (kg/day)

Mass fraction of solid: (0.1155)(2930.93) + (0)M6 = (0.9535-x)M7 (0.9535-x) M7 = 454.29

(kg/day) (kg/day)

Comparing both equations: x = 0.811 Thus, Mass fraction of water = 0.811 Mass fraction of solid = 0.143 M7 = 3196.56 (kg/day) M6 = 265.63 (kg/day) 29

Later on, the outlet of the first mixture is sent to homogenizer. However, in this report, the mass fraction in homogenizer is assumed to be the same because homogenizer is needed only to break the large globules into smaller globules to increase the viscosity of the milk. After the homogenization, not including the pasteurization and cooling stage, the same milk composition is sent to fermenter. At the fermenter, there are a few assumptions which are: 1. The system is assumed to be open system even though it is a semi-batch tank. 2. It is at steady-state. 3. The energy is conserved. 4. The mass is conserved even though the textures are different from the milk. (more jellylike structure produced) 5. Bacteria culture is assumed to be inoculated with NFDM (Non-fat Dry Milk) and the total mass composition in the end of fermentation is 3% of total mass. 6. Assuming the reaction of lactose to lactic acid is conserved and it mass fractions at the outlet is proportional to its reaction. By taking in measure of all the assumptions, earlier, among the solid composition of milk, lactose is also present in the milk about 4.8%. While at the inlet stream of the fermenter now, not including the 4.8% of lactose composition, the other solid composition total is 14.1%. At the end of the fermentation process, 95% of the lactose will be converted into lactic acid. F-101 M8 = ________ kg/day XBacteia (with NFDM) = 1 M9 = _________ kg/day Xwater = x XL.acid = 0.0456 Xlactose = 0.0024 XBacteria(with NFDM) = 0.03 Xother = (1-x-0.0456-0.0024-0.03)

M7 = 3196.56 kg/day Xwater = 0.811 Xlactose = 0.048 Xother = 0.141

FERMENTER 30

M7 + M8 = M9 3196.56 + M8 = M9

(kg/day) (kg/day)

Mass fraction of water: (0.811)(3196.56) + (0)M8 = (x)M9 xM9 = 2592.41

(kg/day) (kg/day)

Mass fraction of others composition: (0.141)(3196.56) + (0)M8 = (0.922-x)M9 (0.922-x)M9 = 450.71

(kg/day) (kg/day)

By comparing the equations: x= 0.785 Thus, Mass fraction of water = 0.785 Mass fraction of others = 0.137 M9 = 3302.43 (kg/day) M8 = 105.87 (kg/day)

After cooling, the outlet from the fermenter stream will be sent to mixer again for addition of stabilizers and sweeteners. In the second mixing, again, proline is used as stabilizer and the sweetener is aspartame. The mass fraction of both of the stabilizer and sweetener is assumed to be 5% of the total end product mass and assumed to be 2.5% each. At the first mixture, proline of total mass 0.5% had already present, thus, by simple calculations, the mass fraction of proline needed is calculated. The bacteria inoculated with NFDM is considered to be solid mass composition at the mixer.

31

M-102 M10 = _________ kg/day Xproline = 0.44 Xaspartame = 0.56

M11 = _______ kg/day Xwater_11 = (x) XL.acid_11 = (x2) Xproline_11 = 0.025 Xaspartame_11 = 0.025 Xsolid_11 = (0.95-x-x2)

M9 = 3302.43 kg/day Xwater = 0.785 XL.acid = 0.0456 Xsolid = 0.1644 Xproline = 0.005

MIXER

M9 + M10 = M11 3302.43 + M10 = M11

(kg/day) (kg/day)

Mass fraction of water: (0.785)(3302.43) + (0)M10 = (x)M11 xM11 = 2592.41

(kg/day) (kg/day)

Mass fraction of lactic acid: (0.0456)(3302.43) + (0)M10 = (x2)M11 x2M11 = 150.59

(kg/day) (kg/day)

Mass fraction of solid: (0.1644)(3302.43) + (0)M10 = (0.95-x-x2)M11 (0.95-x-x2)M11 = 542.92

(kg/day) (kg/day)

32

Comparing the three equations: x2 = 0.044 M11 = 3422.5 kg/day x = 0.757 Thus, mass fraction of water = 0.757 mass fraction of solid = 0.149 mass fraction of lactic acid = 0.044 M10 = 120.07

33

Energy balance Sample calculation

M4 = 66.07 kg/day Xfat_2 = 0.58 Xwhey_2 = 0.42 (l, 50˚C, 1 atm) M5 = 2930.93 kg/day Xwater_5 = 0.8845 Xsolid_5 = 0.083 Xfat_5 = 0.0325 Xwhey_5 = 0 (l, 50˚C, 1 atm)

M3 = 2997 kg/day XSolid_3 = 0.081 Xwhey_3 = 0.009 XWater_3 = 0.865 Xfat_3 = 0.045 (l, 4˚C, 1 atm) CENTRIFUGER

 Inlet stream  Solid Mass solid = 0.081 x 2997 = 242.76 kg/day Converting unit kg/day to kmol/day, since 59% of solid content is lactose, so we assumed mw solid = mw lactose = 342.3 kg/kmol Mass solid = 242.76 kg/day ÷ 342.3 kg/kmol = 0.709 kmol/day 

Whey Mass whey = 0.009 x 2997 = 26.97 kg/day Converting unit kg/day to kmol/day, since 58% of whey content is blactoglobulin, we assumed mw whey = mw b-lactoglobulin = 18300 kg/kmol Mass whey = 26.97 kg/day ÷ 18300 kg/kmol = 0.001 kmol/day 34



Water Mass water = 0.865 x 2997 = 2592.41 kg/day Converting kg/day to kmol/day, since mw water = 18.016 kg/kmol Mass water = 2592.41 kg/day ÷ 18.016 kg/kmol = 143.89 kmol/day



Fat Mass fat = 0.045 x 2997 = 134.87 kg/day Converting kg/day to kmol/day, since mw fat = 891.49 kg/kmol Mass fat = 134.87 kg/day ÷ 891.49 kg/kmol = 0.15 kmol/day

 Oulet stream 1 

Fat Mass fat = 0.58 x 66.07 = 38.32 kg/day Converting kg/day to kmol/day, since mw fat = 891.49 kg/kmol Mass fat = 38.32 kg/day ÷ 891.49 kg/kmol = 0.04 kmol/day



Whey Mass whey = 27.275 kg/day Converting kg/day to kmol/day. Since 58% of whey content is b-lactoglobulin, we assumed mw whey = mw b-lactoglobulin = 18300 kg/kmol Mass whey = 27.275 kg/day ÷ 18300 kg/kmol = 0.002 kmol/day

 Oulet stream 2 

Solid Mass solid = 0.083 x 2930.93 = 243.27 kg/day Converting unit kg/day to kmol/day, since 59% of solid content is lactose, so we assumed mw solid = mw lactose = 342.3 kg/kmol 35

Mass solid = 243.27 kg/day ÷ 342.3 kg/kmol = 0.71 kmol/day 

Water Mass water = 0.8845 x 2930.93 = 2592.41 kg/day Converting kg/day to kmol/day, since mw water = 18.016 kg/kmol Mass water = 2592.41 kg/day ÷ 18.016 kg/kmol = 143.89 kmol/day



Fat Mass fat = 0.0325 x 2930.93 = 95.26 kg/day Converting kg/day to kmol/day, since mw fat = 891.49 kg/kmol Mass fat = 95.26 kg/day ÷ 891.49 kg/kmol = 0.107 kmol/day

Refererence; Solid, whey, water, fat (l, 4˚C, 1 atm) Substance

nin(kmol/day)

Hin(kJ/kmol)

nout(kmol/day)

Hout(kJ/kmol)

Solid

0.709

0

0.700

ΔH1

Water

143.890

0

143.890

ΔH2

Fat

0.150

0

0.107

ΔH3

0.043

ΔH4

0.001

ΔH5

Whey

0.001

0

 Solid Since 59% of solid content is lactose, so we assumed that properties of solid = properties of lactose, which formula for lactose is C12H22O11 Cp solid = cp lactose = 12(12) + 22(18) + 11(25) = 815 kJ/kg ˚C 36

Since unit needed is kJ/kmol ˚C, so that value must be times with mw. Mw lactose = 342.3 kg/kmol Cp solid = 815 kJ/kg ˚C x 342.3 kg/kmol = 278974.5 kJ/kmol ˚C 50

ΔH1 = ∫4 𝑐𝑝 𝑑𝑡 = 13948725 – 1115898 = 12832827 kJ/kmol  Water Cp water = 75.4 kJ/kmol ˚C 50

ΔH2 = ∫4 𝑐𝑝 𝑑𝑡 = 3770 – 301.6 = 3468.4 kJ/kmol  Fat Cp fat = 2.177 kJ/kg K Since the unit needed is kJ/kmol ˚C, that value must be times with mw fat(891.49 kg/kmol) and unit conversion of temperature(274.15 K = 1 ˚C) Cp = 2.1777 kJ/kg K x 274.15 K/˚C x 891.49 kg/kmol = 532059.06 kJ/kmol ˚C 50

ΔH3 = ∫4 𝑐𝑝 𝑑𝑡 = 26602953 – 2128236.24 = 24474716.76 kJ/kmol Therefore ΔH3 = ΔH4  Whey Cp whey = 0.06 kJ/kmol K Since unit needed is kJ/kmol ˚C, the value must be times with unit conversion of temperature( 274.15 K = 1 ˚C) Cp = 0.06 kJ/kmol K x 274.15 K/˚C = 16.45 kJ/kmol ˚C 50

ΔH5 = ∫4 𝑐𝑝 𝑑𝑡 = 822.5 – 65.8 = 756.7 kJ/kmol 37

ΔH = ∑noutHout - ∑ninHin = [(0.700 x 12832827) + (143.890 x 3468.4) + (0.107 x 24474716.76) + (0.043 x 24474716.76) + (0.001 x 756.7)] - 0 = 5.07 x 106 kJ/day Since unit needed is kJ/s, the value must be time with unit conversion of time( 1 day/ 86400 s) ΔH = 5.07 x 106 kJ/day ÷ 86400 s/day = 58.68 kJ/s → 58.68 kw Therefore Q + Ws = ΔH + ΔEk + Δ Ep, Since Ws, ΔEk and ΔEp = 0 Q = ΔH = 58.68 kw

38

2.4.3 Heat Exchanger 2.4.3.1 Heat Transfer Mode, Type flow and Calculations Table 11: Heat transfer properties at heat exchanger

Heat Exchanger E-101 E-102 E-103 E-104

Re

Nu

OHTC

Rf

∆Tlm

1913.76 695.29 191 331.4

35 30.7 53.27 28.78

2.02 1.798 0.0645 21.2

0.00088 0.00053 0.00053 0.00053

9.28 17.24 11.2 8.03

E-101 Assumptions: 1. Average constant thermal properties (thermal conductivity and specific heat) and convective heat transfer coefficient along the heat exchanger. 2. Negligible internal heat generation and negligible free convection.

Tmilk, in = 75.7 °C Twater, out = 65°C

Plate Heat

Tmilk, out = 92°C Twater, in = 100°C

Exchanger

The mode of heat transfer in this tank is convection. Convection refers to heat transfer that occurs between a surface and a moving fluid as a temperature gradient exists. The faster the motion of the fluids, more amount of heat transfer that occurs. There are two types of convection, internal forced convection and external forced convection. (Frank P. Incropera) The heat exchanger that is most commonly used in dairy production is: 1. Plate heat exchanger. 2. Tubular heat exchanger. The heat exchanger that we chose is the plate heat exchanger. This is because the plate exchanger is more widely used in most existing process. Also, advantages of this type of heat 39

exchanger is it offers a large transfer surface that is readily accessible for cleaning (R. L EARLE), superior heat exchanger performance, lower temperature gradient, higher turbulence and compactness over tubular heat exchanger( (Bipan Bansal). Plate heat exchanger is very popular for low viscosity fluid; in this case, the fluid is milk. The plate heat exchanger consists of a stack of corrugated stainless steel plates clamped together in a frame. Heating and cooling fluid flow through alternate tortuous passages between vertical plates. The flow of the fluid is counter current. A counter current heat exchanger is more efficient as it takes a smaller heat transfer to the surface area (As) to achieve the same heat transfer rate (q) as a parallel flow heat exchanger (APPLIED PHYSICS). A counter flow heat exchanger supplies hotter portion of the two fluids at one end, and cold portion at the other end. The temperature distribution for a counter flow heat exchanger is as shown below.

Figure E7.1 Temperature distribution of a counter flow heat exchanger. Reynolds number. To calculate the Reynolds number, the formula used is; 𝑅𝑒 =

𝜌𝑈𝑚𝑥 µ

Where 𝜌 is the density of heating agent, Um is the mean velocity, x is the length and µ is the dynamic viscosity of heating agent. The heating agent used is water. The value of density of water is known to be 1000kg/m3. The value of Um is then calculated by the formula; 𝑈𝑚 =

ṁ 𝜌𝐴

Um is the mean velocity, ṁ is the mass flow rate, 𝜌 is the density of cooling water and A is the cross sectional area. The area, Ac, is calculated by the formula; 40

𝐴𝑐 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 𝑤𝑖𝑑𝑡ℎ Ac = 0.04 m2 (Funke) By knowing this, we can calculate the value of mean velocity, Um 𝑈𝑚 =

0.03599 957.9(0.04)

𝑈𝑚 = 9.39 𝑥 10^ − 4 Inserting the values into the Reynolds formula, 957.9𝑘𝑔 𝑚3 ) 9.39 𝑥 10^ − 4(0.6) 𝑅𝑒 = 0.282 𝑥 10 − 3 (

𝑅𝑒 = 1913.76

The flow is laminar as it is less than 2300.

Nusselt number Nu = 0.664 Rex1/2Pr1/3 By inserting the value; Nu = 0.664 (1913.76)1/2(1.75)1/3 Nu = 35

Tlm The formula used to calculate Tlm is; 𝛥𝑇𝑙𝑚 =

(𝑇ℎ𝑖 − 𝑇𝑐𝑜) − (𝑇ℎ𝑜 − 𝑇𝑐𝑖) 𝑇ℎ𝑖 − 𝑇𝑐𝑜 ln[ ] 𝑇ℎ𝑜 − 𝑇𝑐𝑖

41

However, the outlet temperature of the heating agent is still unknown. This temperature can be obtained by calculating the qc of the milk. Since the qh and qc is the same, therefore, the outlet temperature of water can be calculated. By using the formula; 𝑞𝑐 = ṁ𝐶𝑝𝑐 (𝑇𝑐, 𝑜 − 𝑇𝑐, 𝑖) Where ṁ is the mass flow rate, Cp is the specific heat capacity of the water, Th,i is the temperature inlet of cold fluid and Th.o is the temperature outlet of the cold fluid. Cp of the fluid is known to be 3.77 kJ/kg.°C (The Engineering Toolbox). Ṁ is calculated to be 3196.56 kg/day. Through unit conversion, the ṁ in kg/s is 0.03537 kg/s. The ΔT is 27°C. Thus the q is 3.76kJ/s. 3196.56 kg day

1 day

1 hr

1min = 0.03699kg/s

24hr

60min 60s

Since the qc is equal to qh, therefore 𝑞ℎ = ṁ𝐶𝑝ℎ (𝑇ℎ, 𝑖 − 𝑇ℎ, 𝑜) 3.76 = 0.03699(4.187) (100 − 𝑇ℎ, 𝑜) 𝑇ℎ, 𝑜 = 75.7°𝐶 Thus, the ΔTlm can be calculated 𝛥𝑇𝑙𝑚 =

(100 − 92) − (75.7 − 65) 100 − 92 ln[ ] 75.7 − 65

𝛥𝑇𝑙𝑚 = 9.2846 °𝐶 Overall heat transfer coefficient (OHTC) To find the value of the overall heat transfer coefficient, U, the formula used is: 𝑈=

𝑞 𝑇𝑙𝑚. 𝐴

Where U is the overall heat transfer coefficient, q is the rate of heat transfer and A is the surface area. The surface area As can be calculated by using the formula 𝐴𝑠 = 𝑁𝒙 𝒍𝒆𝒏𝒈𝒕𝒉 𝒙 𝒘𝒊𝒅𝒕𝒉 𝐴𝑠 = 5 𝑥 0.04 𝐴𝑠 = 𝟎. 𝟐 𝒎^𝟐 42

By inserting the values, 𝑈=

3.76 9.2846(0.2)

𝑈 = 2.0248 𝑘𝑊/𝑚2°C Fouling factor The Fouling factor for water above 50°C is 0.00088m2K/W (Engineering page)

E-102 Assumptions: 1. Average constant thermal properties (thermal conductivity and specific heat) and convective heat transfer coefficient along the heat exchanger. 2. Negligible internal heat generation and negligible free convection.

Tmilk, in = 92°C

Twater, out = 72.29°C

Plate

Tmilk, out =45°C

Heat Twater, in =30°C Exchanger

The type of heat exchanger used is the plate heat exchanger. This is because a plate heat exchanger is known to effectively handle low viscosity fluids. The plate heat exchanger consists of a stack of corrugated stainless steel plates clamped together in a frame. Heating and cooling fluid flow through alternate tortuous passages between vertical plates. Type of flow in this heat exchanger is the counter current flow. A counter flow heat exchanger supplies hotter portion of the two fluids at one end, and cold portion at the other end. Thus, heat transfer occurs between the hotter portions of the two fluids at one end (Frank P. Incropera). The mode of heat transfer involved is convection. Convection occurs as the fluid is in motion and there is a bounding surface when the two are at different temperatures.

43

Reynolds number The formula used to calculate Reynolds number is 𝑅𝑒 =

𝜌𝑈𝑚𝑥 µ

Where 𝜌 is the density of the working fluid which is water, Um is the mean velocity, x is the length and µ is the dynamic viscosity of the water. The mean velocity is calculated by using the formula 𝑈𝑚 =

ṁ 𝜌𝐴𝑐

To calculate the cross sectional area, 𝐴 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 𝑤𝑖𝑑𝑡ℎ A = 0.04 By knowing this, it is possible to calculate the value of Um. 𝑈𝑚 =

0.03699 996.0(0.04)

𝑈𝑚 = 9.2846 𝑥 10^ − 4 Thus, the Reynolds number can be calculated as 𝑅𝑒 =

996.0(9.2846 𝑥 10−4 )0.6 0.798 𝑥 10^3 𝑅𝑒 = 695.29

Since the Re <2300, the flow is laminar. Nusselt number The nusselt number formula is given by Nu = 0.664 Rex1/2Pr1/3 Nu = 0.664 (695.29)1/2(5.42)1/3 Nu = 30.7

44

Tlm (𝑇ℎ𝑖 − 𝑇𝑐𝑜) − (𝑇ℎ𝑜 − 𝑇𝑐𝑖) 𝑇ℎ𝑖 − 𝑇𝑐𝑜 ln[ ] 𝑇ℎ𝑜 − 𝑇𝑐𝑖

𝛥𝑇𝑙𝑚 =

Where Tlm is T log mean, Th,i and Th,o is the temperature hot inlet and outlet respectively and Tc,o and Tc,i are the outler and inlet temperature of the cold fluid which is the working fluid. Since the Tc,o is unknown, it can be calculated by determining the value of qh 𝑞ℎ = ṁ𝐶𝑝ℎ (𝑇ℎ, 𝑖 − 𝑇ℎ, 𝑜) 𝑞ℎ = (0.03699)(3.77) (92 − 45) 𝑞ℎ = 6.55𝑘𝐽/𝑠 Since qh = qc, 𝑞𝑐 = ṁ𝐶𝑝 (𝑇𝑐, 𝑜 − 𝑇𝑐, 𝑖) 6.55 = (0.03699)(4.187) (𝑇𝑐, 𝑜 − 30) 𝑇𝑐, 𝑜 = 72.29 °𝐶 Thus, the ΔTlm can be calculated by substituting in the values. 𝛥𝑇𝑙𝑚 =

(𝑇ℎ𝑖 − 𝑇𝑐𝑜) − (𝑇ℎ, 𝑜 − 𝑇𝑐𝑖) 𝑇ℎ𝑖 − 𝑇𝑐𝑜 ln[ ] 𝑇ℎ𝑜 − 𝑇𝑐𝑖

𝛥𝑇𝑙𝑚 =

(92 − 72.29) − (45 − 30) 92 − 72.29 ln[ ] 45 − 30 𝛥𝑇𝑙𝑚 = 17.24 °𝐶

OHTC By using the LMTD method, 𝑈=

𝑞 𝑇𝑙𝑚. 𝐴

Where U is the overall heat transfer coefficient, q is equal to rate of heat transfer, Tlm is the T log mean and A is the surface area. 45

By inserting the value, 𝑈=

6.2 (17.24)(0.2)

𝑈 = 1.798 𝑘𝑊/𝑚2°C

Fouling factor The Fouling factor for temperature below 50°C is 0.00053m2K/W (Engineering page)

E-103 Assumptions: 1. Average constant thermal properties (thermal conductivity and specific heat) and convective heat transfer coefficient along the heat exchanger. 2. Negligible internal heat generation and negligible free convection.

Tmilk, in = 46°C Twater, out = 34°C

Tmilk, out = 30°C

Tubular Heat

Twater, in = 20°C

Exchanger

The type of heat transfer used is turbular heat exchanger. The reason that we chose this type of heat exchanger is because that it is cheaper than the plate heat exchanger. The type of flow used in this heat exchanger is the counterflow. A counterflow heat exchanger has the hot fluid entering at one end of the heat exchanger flow path and the cold fluid entering at the other end of the flow path (Bengston).

Reynolds number The Reynolds formula for turbular heat exchanger is the same as plate heat exchanger, that is 𝑅𝑒 =

𝜌𝑈𝑚𝐷 µ 46

Where Re is the Reynolds number, 𝜌 is the density of fluid, Um is the mean velocity, D is the diameter of the tube and µ is the dynamic viscosity of the fluid. To calculate the re, we must first calculate the Um. 𝑈𝑚 =

ṁ 𝜌𝐴𝑐

Um is the mean velocity, ṁ is the mass flow, 𝜌 is fluid density, and Ac is the cross sectional area. To calculate the cross sectional area, 𝐴𝑐 = 𝞹𝒓2 Ac = (0.0635)2 Ac = 0.01266m2

The mass flow rate must be converted to kg/s before doing further calculation. 3302.43 kg day

1 day 1hr 24hr

1min = 0.03822 kg/s

60min 60 sec

Thus, the Um is calculated to be 𝑈𝑚 = 𝑈𝑚 =

ṁ 𝜌𝐴𝑐

0.03822 998.0(0.01266)

𝑈𝑚 = 3.02 𝑥 10^ − 3 Then, the Reynolds number can be calculated by 𝑅𝑒 =

(998.0)(3.02 𝑥 10^ − 3)(0.0635) 1.002 𝑥 10^ − 3 𝑅𝑒 = 191

The flow is laminar as the Re is 180 which is less than 2300. Nusselt number D 0.065 ( L ) Re Pr Nu = 3.66 + D 1 + 0.04 [( L ) Re Pr]^2/3 47

0.0635 ) 191( 7.01) 0.5 Nu = 3.66 + 0.0635 1 + 0.04 [( ) (191) (7.01)]^2/3 0.5 0.065 (

Nu = 53.27

Tlm 𝛥𝑇𝑙𝑚 =

(𝑇ℎ𝑖 − 𝑇𝑐𝑜) − (𝑇ℎ𝑜 − 𝑇𝑐𝑖) 𝑇ℎ𝑖 − 𝑇𝑐𝑜 ln[ ] 𝑇ℎ𝑜 − 𝑇𝑐𝑖

Where Th,i and Th,o is the temperature hot inlet and outlet and Tc,i and Tc,o is the temperature cold inlet and outlet. 𝑞ℎ = ṁ𝐶𝑝ℎ(𝑇ℎ, 𝑖 − 𝑇ℎ, 𝑜) 𝑞ℎ = 0.03822(3.77)(45 − 30) 𝑞ℎ = 2.16 𝑊 𝑞𝑐 = ṁ𝐶𝑝𝑐(𝑇𝑐, 𝑜 − 𝑇𝑐, 𝑖) 2.16 = 0.03822(4.187)(𝑇𝑐, 𝑜 − 20) 𝑇𝑐, 𝑜 = 33.5°𝐶 Thus, 𝛥𝑇𝑙𝑚 =

(𝑇ℎ𝑖 − 𝑇𝑐𝑜) − (𝑇ℎ𝑜 − 𝑇𝑐𝑖) 𝑇ℎ𝑖 − 𝑇𝑐𝑜 ln[ ] 𝑇ℎ𝑜 − 𝑇𝑐𝑖

𝛥𝑇𝑙𝑚 =

(46 − 33.5) − (30 − 20) 46 − 33.5 ln[ 30 − 20 ] 𝛥𝑇𝑙𝑚 = 11.2°𝐶

48

OHTC 𝑈=

𝑞 𝑇𝑙𝑚. 𝐴

Where U is the overall heat transfer coefficient, q is equal to rate of heat transfer, Tlm is the T log mean and A is the surface area. To find As, 𝐴𝑠 = 𝑁𝞹𝑫𝑳 𝐴𝑠 = (30)𝞹(𝟎. 𝟎𝟔𝟑𝟓)(𝟎. 𝟓) 𝐴𝑠 = 2.99𝑚2 To find the value of U, 𝑈=

2.16 (11.2)(2.99) 𝑈 = 0.0645𝑘𝑊/𝑚2°C

Fouling factor The Fouling factor for temperature below 50°C is 0.00053m2K/W (Engineering page) E-104 Assumptions: 1. Average constant thermal properties (thermal conductivity and specific heat) and convective heat transfer coefficient along the heat exchanger. 2. Negligible internal heat generation and negligible free convection.

Tmilk, in = 30°C

Twater, out = 17.88°C

Plate Heat

Tmilk, out = 5°C

Twater, out =0.01°C

Exchanger

The mode of heat transfer used in this heat exchanger is the convection. This is because the fluid is moving in a boundary with different temperatures. 49

The type of heat exchanger used is the plate heat exchanger. This is because we need to cool the fluid to a very low temperature which is 5°C from 30°C. The working fluid used is water. Water is chosen as it can be recycled back into the streams and it is easily available. The plate heat exchanger consists of a stack of corrugated stainless steel plates clamped together in a frame. Heating and cooling fluid flow through alternate tortuous passages between vertical plates. The flow is counterflow. . A counterflow heat exchanger has the hot fluid entering at one end of the heat exchanger flow path and the cold fluid entering at the other end of the flow path (Bengston). Reynolds number The Reynolds number is calculated by 𝑅𝑒 =

𝜌𝑈𝑚𝐷 µ

Where Re is the Reynolds number, 𝜌 is the density of cooling fluid, Um is the mean velocity, D is the diameter of tube and µ is the dynamic viscosity of the working fluid. In this heat exchanger, the cooling or known as the working fluid is water. Water is chosen as it is easily obtained and more cost friendly. To calculate Um, the formula is 𝑈𝑚 =

ṁ 𝜌𝐴𝑐

Where ṁ is the mass flow rate, Um is the mean velocity, 𝜌 is the density of fluid and Ac is the cross sectional area. By calculations, we have known that the Ac is equal to the 0.04227m2.The mass flow rate however, is different and calculations are need to be done to converse the unit to kg/s from kg/hr 3422.5 kg day

1day

1hr

1min = 0.0396 kg/s

24hr

60min 60s

Thus, 𝑈𝑚 =

0.0396 999.8(0.04)

𝑈𝑚 = 9.90 𝑥 10^ − 4

50

𝑅𝑒 =

999.8(9.90 𝑥 10^ − 4)(0.6) 1.792 𝑥 10^ − 3 𝑅𝑒 = 331.4

The flow is laminar as the Reynolds number is less than 2300. Nusselt number Nu = 0.664 Rex1/2Pr1/3 Nu = 0.664 (331.4)1/2(13.5)1/3 Nu = 28.78

Tlm 𝛥𝑇𝑙𝑚 =

(𝑇ℎ𝑖 − 𝑇𝑐𝑜) − (𝑇ℎ𝑜 − 𝑇𝑐𝑖) 𝑇ℎ𝑖 − 𝑇𝑐𝑜 ln[ ] 𝑇ℎ𝑜 − 𝑇𝑐𝑖

Where Th,i and Th,o is the temperature hot inlet and outlet and Tc,i and Tc,o is the temperature cold inlet and outlet. 𝑞ℎ = ṁ𝐶𝑝ℎ(𝑇ℎ, 𝑖 − 𝑇ℎ, 𝑜) 𝑞ℎ = 0.0396 (3.77)(30 − 5) 𝑞ℎ = 3.4056𝑊 𝑞𝑐 = ṁ𝐶𝑝(𝑇𝑐, 𝑜 − 𝑇𝑐, 𝑖) 3.4056 = 0.0396(4.81)(𝑇𝑐, 𝑜 − 0.01) 𝑇𝑐, 𝑜 = 17.88°𝐶 Thus, Tlm 𝛥𝑇𝑙𝑚 =

(30 − 17.88) − (5 − 0.01) 30 − 17.88 ln[ ] 5 − 0.01 𝛥𝑇𝑙𝑚 = 8.03°𝐶

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OHTC 𝑈=

𝑞 𝑇𝑙𝑚. 𝐴

Where U is the overall heat transfer coefficient, q is equal to rate of heat transfer, Tlm is the T log mean and A is the surface area. To find the value of U, 𝑈=

3.4056 (8.03)(0.02) 𝑈 = 21.2 𝑘𝑊/𝑚2°C

Fouling factor The Fouling factor for temperature below 50°C is 0.00053m2K/W (Engineering page)

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2.5

Bioprocess and Metabolic Regulations 2.5.1 Biomolecules Involved

Bio-molecules are organic compounds that are essential to living organisms (Biomolecules, 2011). Bio-molecules are derived from the simplest organic molecule that is hydrocarbon (McKee, 2003). The type of bio-molecules involved are from carbohydrate and protein biomolecules. Lactose, glucose and galactose are part of carbohydrate bio-molecule while lactase is part of protein bio-molecule.

2.5.1.1 Lactose Lactose is a dissacharide with molecular formula of C12H12O11. Lactose is a some kind of sugar found in milk (Calvero, 2013). Lactose is composed of glucose and galactose that is linked via β1,4-glycosidic bond. The linkage occurs between hydroxyl group of carbon 1 of galactose and hydroxyl group of carbon 4 of glucose. Lactose is a reducing sugar because glucose component consists a hemiacetal group (McKee, 2003).

Figure 5: Chemical structure of lactose (Calvero, 2013)

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Lactose is also known as 4-O-β-D-galactosylpyranosyl-α-D-glucopyranoside. This is the IUPAC name for lactose. Lactose comes from the cattle’s milk that is used in our production of yoghurt. Lactose is catalyzed by enzyme lactase provided by the bacteria when the milk undergoes fermentation. From lactose anabolic pathway, galactose and glucose are formed (H.Garret & Grisham, 2010).

2.5.1.2 Glucose Glucose is a carbohydrate and it is called a monosaccharide which is a simple sugar. Glucose it also known as dextrose with a molecular formula of C6H12O6 or H-(C=O)-(CHOH)5-H (Nave, 2012).

A monosaccharide has a ring with a hemiacetal functional group (Monosaccharide-

Structure of Glucose, 2001). Six-membered hemiacetal rings are called pyranoses as it is similar to pyran. In pyranose form, glucose is known as glucopyranose (McKee, 2003). Glucose is primarily from corn syrup and it serves as energy sources for plants and animals (Nave, 2012). Glucose is needed in glycolysis pathway in order to produce pyruvate which will then be used in lactic acid fermentation (McKee, 2003).

Figure 6: Chemical structure of glucose (Nave, 2012)

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Figure 7: Hemiacetal functional group in glucose (Monosaccharide-Structure of Glucose, 2001)

2.5.1.3 Galactose Galactose is a monosaccharide of carbohydrate which is known as simple sugar. It is an aldose, hexose and also a reducing sugar. It is called a reducing sugar because it contain aldehyde functional group. The molecular formula for galactose is the same as glucose that is C6H12O6. The difference between glucose and galactose is the position of 4th carbon in molecular structure. The –OH group at 4th carbon of galactose in the upward projection in the chair form compared to glucose which is in horizontal projection in the chair form. Galactose also has hemiacetal goup. A carbon which has both ether oxygen and alcohol functional group is hemiacetal and known as anomeric carbon (Ophardt, Galactose, 2003).

55

Figure 8: Molecular structure of galactose (Ophardt, Galactose, 2003)

Figure 9: Difference between galactose and glucose in structure (Ophardt, Galactose, 2003)

Galactose produced will be converted to glucose-6-phosphate to undergo glycolysis to formed pyruvate. Pyruvate will then be converted into lactic acid.

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2.5.1.4 Lactase Lactase is an enzyme made out amino acids. All enzymes are proteins. Thus, lactase is part of protein bio-molecule. Lactase is used to cleave the glycosidic bond in lactose to form glucose and galactose (Carroll, 2013)

2.5.2 Biochemical Pathway In production of yoghurt from bacteria, Lactobacillus bulgaricus and Streptococcus thermophillus grow vigorously on milk’s lactose (H.Garret & Grisham, 2010). Lactic acid fermentation occurs when lactose, a sugar which is composed of glucose and galactose is converted to lactate. Lactase is an enzyme provided by bacteria which catalyze the reaction of converting lactose to glucose and galactose.

Figure 10: Conversion of lactose to galactose and glucose (Taylor & Stahlberg, 2005)

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Figure 11: Overview of glycolysis (Glycolysis, 2013)

Glucose that is produced from lactose catalyzation enters gycolysis pathway. Glycolysis is a process or pathway where glucose is converted to three-carbon pyruvate and it is viewed in ten steps of reaction. It occurs without the presence of oxygen which is known as anaerobic metabolism (H.Garret & Grisham, 2010). C6H12O6 + 2 NAD+ + 2 ADP + 2 P -----> 2 pyruvic acid, (CH3(C=O)COOH + 2 ATP + 2NADH + 2 H+ Equation 3: Overall reaction of glycolysis (Ophardt, Glycolysis Summary, 2003)

Glycolysis is divided into two phases that is first phase and second phase. In first phase, glucose is converted to two molecules of glyceraldehyde-3-phosphate and during second phase, two molecules of pyruvate are produced (H.Garret & Grisham, 2010).

58

For first phase, there are five reactions. During this phase, energy is used in order to gain more. Reaction 1 is known as phosphorylation of glucose, a six-carbon atom. Glucose is converted to glucose-6-phosphate by enzyme hexokinase or glucokinase. Enzyme hexokinase is used to phosphorylate glucose and stored it in cell. This enzyme is regulated and is allosterically inhibited by glucose-6-phosphate. Adenine triphosphate, ATP is consumed in this reaction (H.Garret & Grisham, 2010).

Figure 12: Phosphorylation of glucose (Helmenstine, 2013)

Next is 2nd reaction. It occurs when phosphoglucoisomerase catalyzes the isomerization of glucose-6-phosphate. Aldose glucose-6-phosphate converts to ketose fructose-6-phosphate by enzyme phosphoglucoseisomerase (H.Garret & Grisham, 2010).

Figure 13: Conversion of glucose-6-phosphate to fructose-6-phosphate (Helmenstine, 2013)

Second phosphorylation happens in reaction 3. This phosphorylation is driven by ATP where it is consumed again for the second time in this step. Phosphofructokinase-1 catalyzes the changes of

fructose-6-phosphate

to

fructose-1,6-biphosphate.

In

glycolysis

pathway, 59

phosphofructokinase is the major regulatory enzyme and its activity is allosterically inhibited by citrate and high levels ATP (McKee, 2003).

Figure 14: Phosphorylation of fructose-6-phosphate (Helmenstine, 2013)

Phase 1 of glycolysis ends with the cleavage of fructose-1,6-bipshosphate into two three-carbon molecules. Enzyme fructose biphosphate aldolase catalyzes fructose-1,6-biphosphate to form dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P) (McKee, 2003).

Figure 15: Cleavage of fructose-1,6-phosphate (Helmenstine, 2013)

For next reaction, only glyceraldehyde-3-phosphate is needed. Thus, dihdroxyacetone phosphate produced in 4th reaction is converted to glyceraldehydes-3-phosphate by enzyme triose phosphate isomerase. In this 5th reaction, two molecules of glyceraldehyde-3-phosphate are produced (H.Garret & Grisham, 2010).

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Figure 16: Interconversion of glyceraldehaydes-3-phosphate and dihydroxyacteone phosphate

Second phase of glycolysis pathway consists another five series of reactions. There are two processes involved in 6th reaction which are oxidation and phosphorylation of glyceraldehyde-3phosphate. This reaction is catalyzes by glyceraldehyde-3-phosphate dehydrogenase that contain two binding sites for glyceraldehydes-3-phosphate and NAD+ to produce glycerate-1,3biphosphate.

While this reaction is happening, NAD+ (nicotinamide adenine dinucleotide)

undergoes reduction to form NADH (McKee, 2003).

Figure 17: Oxidation of glyceraldehyde-3-phosphate

For 7th reaction, phosphoryl group is transferred and ATP is synthesized when glycerate-1, 3biphosphate is catalyzes by phosphoglycerate kinase to adenosine diphosphate, ADP. Reaction 7 is known as substrate level phosphorylation, this is due to the yielding of ATP caused by transfer of phosphoryl group from substrate (McKee, 2003).

Figure 18: Phosphoryl group transfer

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Then, enzyme phosphoglycerate mutase catalyzes the migration of functional group within the subunit that converts 3-phosphoglcerate to 2-phosphoglycerate (H.Garret & Grisham, 2010).

Figure 19: Interconversion of 3-phosphoglycerate to 2-phosphoglycerate

2-phosphoglycerate is catalyzed by enolase to form phosphoenolpyruvate (PEP). In this 9th reaction, water is removed from 2-phosphoglycerate to form phosphoenolpyruvate’s enol structure (H.Garret & Grisham, 2010).

Figure 20: Dehydration of phosphoenolpyruvate

The last reaction in glycolysis occurs when phosphoenolpyruvate (PEP) is driven by pyruvate kinase to form pyruvate. There is a transfer of phosphoryl group from phosphoenolpyruvate to ADP that synthesis ATP (H.Garret & Grisham, 2010).

Figure 21: Synthesis of pyruvate

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Galactose that is formed from catalyzation of lactose is required to undergo a few reactions to ensure that it can enter glycolysis pathway to form pyruvate.

Figure 22: Galactose metabolism

Galactose is initially transformed into galactose-1-phosphate by galactokinase. Then galactose1-phosphate is converted to nucleotide UDP-galactose by uridyl transferase. UDP-glucose is formed by isomerisation of galactose. This reaction is catalyzed by UDP-glucose-4-epimerase. UDP-glucose is then converted to glucose-1-phosphate by UDP-glucose-pyrophosphorylase. Glucose-6-phosphate enters glycolysis pathway when it is converted from glucose-1-phosphate by enzyme phosphoglucomutase (McKee, 2003). After obtaining pyruvate molecules from glycolysis pathway entered by glucose and galactose, the pyruvate will then be converted to lactate by lactate dehydrogenase. This whole process is known as lactic acid fermentation (H.Garret & Grisham, 2010)

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. Figure 23: Lactic acid fermentation

There are a few assumptions have been made based on biochemical reactions involved. Firstly, in biochemical reactions there are a simple organic reaction mechanism as an enzyme usually does one conversion at one time. Secondly, the number of reactions occurred are large but the type of reactions involved is usually small.

Lastly, the central importance’s reactions in

biochemistry are few as the one used in energy production and also the synthesis and degradation of major cell components.

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CHAPTER THREE: CONCLUSION AND RECOMMENDATIONS To produce yoghurt, there are several processes that had to proceed in order to produce tasty yoghurt.

In this production, yoghurt is produced by bacteria where the bacteria used are

Lactobacillus bulgaricus and Streptococcus thermophillus. The bacteria are cultured in a tank between the ranges of 35-45 degree celcius.

The first process is filtration, followed by

centrifugation and mixing. The mixture undergoes homogenization process after the temperature is increased to alter the heat.

After the homogenization process, it undergoes pasteurization

process to kill the microorganism. The mixture undergoes cooling process to make sure it suits the optimum temperature of the selected lactic acid bacteria during fermentation process. After the fermentation process, the mixture is cooled again. Stabilizers and flavoring are during the second mixing process. Then it is cooled again for the third time. After cooling it to desired temperature for the yoghurt, the yoghurt is now ready to be packaged and stored.

All these

processes are done where all its mass balances, energy balances and overall heat transfer coefficients calculations are calculated. The processes involved do not violate the first or the second laws of thermodynamics. As a conclusion, the production of yoghurt from bacteria procedures is a success.

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