Yogurt Production Our Thesis Complete -1

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

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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 DATE: 10/6/2013

ID : 2012888124

----------------------------------------

NAME: MOHD NAZMIE BIN MOHAMED MOKHATR DATE: 10/6/2013

ID : 2012880126

----------------------------------------

NAME: MUSALMAH BINTI ADANAN DATE: 10/6/2013

ID : 2012218062

----------------------------------------

NAME: HAZIRAH BINTI HAFIZ DATE: 10/6/2013

ID : 2012434468

3

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 :______________________

4

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

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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

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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 Lactic Acid Bacteria

Sp. Species

OHTC Overall Heat Transfer Coefficient

Re Reynolds

Nu Nusselt

NADH Reduce nicotinamide adenine dinucleotide

DHAP Dihydroxyacetone phosphate

NAD Nicotonamide adenine dinucleotide

LIST OF SYMBOLS

Symbol

°C Degree Celcius

α Alpha

β Beta

µ Viscosity

Δ Changes

∑ Summation

Cp Specific Heat capacity at constant Pressure

ΔTlm Temperature log mean

h Enthalphy

Q Heat transfer

W Work

U Internal energy

v Specific volume

P Pressure

T 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

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

acid.

Filtration

Centrifugation

Mixing

Heat Treatment

Homogenization

Pasteurization

Cooling

Fermentation

Cooling

Mixing

Cooling

Packaging and Storage

Stabilizers and Flavoring

Powder Skimmed milk

Stabilizer

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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

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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

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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 Assumptions

5 1. The flow of liquid is steady-flow

2. The filter completely filtered impurities

7 1. Steady-flow process

2. Whey protein and undesired fat composition are completely

removed after centrifugation.

3. The pressure is assumed 1 atm

10 1. The homogenizer is assumed single-phase homogenizer.

2. Steady-flow process

3. The mass is conserved in homogenizer.

14 1. The fermenter is assumed as open system.

2. The energy is conserved in the fermenter due to constant

temperature.

3. The composition is assumed conserved even though the

textures become more jelly-like.

4. Steady-state during the fermentation process

15-19 1. Steady-flow process.

2. The properties of milk and yogurt entering the heat

exchanger are considered the same as water.

3. Heat loss to surrounding is considered negligible.

Equipments Assumption

FL-101 1. Steady-flow process

2. Impurities are completely removed. CF-101 1. Steady-flow process

2. Heat loss to surrounding is negligible

3. Undesired composition is assumed removed. HG-101 1. Steady-flow process

2. Mass is assumed conserved

3. No heat loss to surrounding where it’s negligible. F-101 1. Steady-flow process

2. Assumed as open system.

3. Energy is assumed conserved.

4. Mass is assumed conserved. E-101 E-102 E-103 E-104

1. Steady flow process

2. 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

22

FM-101 Volumetric Flow meter

ST-101 ST-102

Temporary storage tank

FL-101 Filter

CF-101 Centrifuge

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

Heat Exchanger

HG-101 Homogenizer

M-101 M-102

Mixer

CT-101 Culture Tank

F-101 Fermenter

P-101 Pump

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

Stream 1 2 3 4 5

Temperature (◦C) 4 4 4 27 4

Pressure (atm) 1 1 1 1 1

Mass flow (kg/day) 3000 3000 3000 3 2997

Component Raw milk Raw milk Raw milk Impurities Milk

Stream 6 7 8 9 10-12

Temperature (◦C) 50 65 92 45 40


Mass flow (kg/day) 66.07 2930.93 265.63 3196.56 3196.56

Component Undesired

Composition

Milk Proline

skimmed

milk

Concentrated

Milk

Concentrated

milk

(HE stream)

Stream 13 14 15 16 17

Temperature (◦C) 45 30 30 5 32.5


Mass flow (kg/day) 105.87 3302.43 3302.43 3422.5 3422.5

Component Culture inoculated

with NFDM

Raw

Yogurt

Raw

Yogurt

(HE stream)

Yogurt Cooling and

storage

Stream 18-20 21 22&23

Temperature (◦C) 35.2 27 65

Pressure (atm) 1 1 1

Mass flow (kg/day) 122 120.07 121.7749

Component Working fluid Proline

Aspartame

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 ST-101 ST-102

MOC* SS SS

Type Cone roof Cone roof

Component Raw milk Raw milk

Inlet Temperature (◦C) 4 4

Inlet Pressure (atm) 1 1

Mass capacity (kg/day) 3000 3000

Table 4: Equipment table for fermenter

Fermenter F-101

MOC* SS

Type Plug flow

Component Milk mixture and Bulk Culture

Temperature (◦C) 45

Pressure (atm) 1

Volume (m3) 4

Mass capacity (kg/day) 3500

Component Raw milk

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Table 5: Equipment table for filter

Filter FL-101

MOC* SS

Type Nylon-filter

Component Raw milk

Inlet Temperature (◦C) 4

Inlet Pressure (atm) 1

Outlet Pressure (atm) 1

Mass flow in (kg/day) 3000

Mass flow out (kg/day) 2997

Filtrate flux (kg/day) 3

Area (m2) 27.63

Table 6: Equipment table for centrifuge

Centrifuger CF-101

MOC* SS

Type Disc bowl centrufger

Mass capacity (kg/day) 3000

Component Raw milk

Temperature (◦C) 50

Pressure (atm) 1

Revolution per minute (rpm) 7000

Table 7: Equipment table for mixers

Mixers M-101 M-102

MOC* SS SS

Type Closed vessel with

agitator

Closed vessel with agitator

Component Raw milk

Powder skimmed

milk

Stabilizer (Proline)

Raw yogurt

Stabilizer (Proline)

Aspartame

Inlet Temperature (◦C) 50 30

Inlet Pressure (atm) 1 1

Mass capacity (kg/day) 3500 3500

Mixing time (hr) 0.5 0.5

Volume (ft3) 2 2

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Table 8: Equipment table for homogenizer

Homogenizer FM-101

MOC* SS

Type Single stage

Component Mixture of milk

Temperature in (◦C) 50

Temperature out (◦C) 65

Inlet pressure (atm) 1

Pressure (atm) 178

Table 9: Equipment table for storage freezer

Storage freezer SF-101

MOC* SS

Component Yogurt

Inlet temperature (◦C) 64

Outlet temperature (◦C) 37

Pressure (atm) 1

Mass flow (kmol/hr) 121.77

Heat duty (kW) 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.

26

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

M1 = M2 + M3 (kg/day)

3000 = M2 + M3 (kg/day)

Milk mass fraction:

(0.999)(3000) = (0)M2 + (1)M3 (kg/day)

M3 = 2997 (kg/day)

Impurities mass fraction:

(0.001)(3000)= (1)M2 + (0)M3 (kg/day)

M2 = 3 (kg/day

M3 = _________kg/day

Xmilk = 1

Ximpurities = 0

M1 = 3000 kg/day

Xmilk = 0.999

Ximpurities = 0.001

M2 = ________ kg/day

Ximpurities = 1

FILTER

27

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 86.5 Lactose 4.8 Fat 4.5 Whey protein 0.9 Protein 2.6 Other 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

M3 = M4 + M5 (kg/day)

2997 = M4 + M5 (kg/day)

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

M4 = ________ kg/day

Xfat_2 = 0.58

Xwhey_2 = 0.42

CENTRIFUGER

28

Water mass fraction:

(0.865)(2997) = (0)M4 + (x)M5 (kg/day)

(x)M5 = 2592.41 (kg/day)

Solid mass fraction:

(0.081)(2997) = (0)M4 + (0.9675-x)M5 (kg/day)

(0.9675-x)M5 = 242.76 (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 (kg/day)

M4 = 66.07 (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.

29

M-101

M5 + M6 = M7 (kg/day)

2930.93 + M6 = M7 (kg/day)

Mass fraction of water:

(0.8845)(2930.93) + (0)M6 = (x)M7 (kg/day)

xM7 = 2592.41 (kg/day)

Mass fraction of solid:

(0.1155)(2930.93) + (0)M6 = (0.9535-x)M7 (kg/day)

(0.9535-x) M7 = 454.29 (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)

M7 = __________kg/day

Xwater = x

Xs.milk = 0.0415

Xproline = 0.005

Xsolid = (1-x-0.0415-

0.005)

M5 = 2930.93 kg/day

Xwater = 0.8845

Xsolid = 0.1155

M6 =___________ kg/day

Xproline = 0.11

Xs.milk = 0.89

MIXER

30

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 jelly-

like 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

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

M8 = ________ kg/day

XBacteia (with NFDM) = 1

FERMENTER

31

M7 + M8 = M9 (kg/day)

3196.56 + M8 = M9 (kg/day)


(0.811)(3196.56) + (0)M8 = (x)M9 (kg/day)

xM9 = 2592.41 (kg/day)

Mass fraction of others composition:

(0.141)(3196.56) + (0)M8 = (0.922-x)M9 (kg/day)

(0.922-x)M9 = 450.71 (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.

32

M-102

M9 + M10 = M11 (kg/day)

3302.43 + M10 = M11 (kg/day)


(0.785)(3302.43) + (0)M10 = (x)M11 (kg/day)

xM11 = 2592.41 (kg/day)

Mass fraction of lactic acid:

(0.0456)(3302.43) + (0)M10 = (x2)M11 (kg/day)

x2M11 = 150.59 (kg/day)

Mass fraction of solid:

(0.1644)(3302.43) + (0)M10 = (0.95-x-x2)M11 (kg/day)

(0.95-x-x2)M11 = 542.92 (kg/day)

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

M10 = _________ kg/day

Xproline = 0.44

Xaspartame = 0.56

MIXER

33

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

34

Energy balance

Sample calculation

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 b-

lactoglobulin, we assumed mw whey = mw b-lactoglobulin = 18300 kg/kmol

Mass whey = 26.97 kg/day ÷ 18300 kg/kmol

= 0.001 kmol/day

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)

M4 = 66.07 kg/day

Xfat_2 = 0.58

Xwhey_2 = 0.42

(l, 50˚C, 1 atm)

CENTRIFUGER

35

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



= 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

36

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



= 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

Whey 0.001 0 0.001 ΔH5

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

37

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

ΔH1 = ∫ 𝑐𝑝 𝑑𝑡50

4

= 13948725 – 1115898

= 12832827 kJ/kmol

Water

Cp water = 75.4 kJ/kmol ˚C


4

= 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


4

= 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


4

= 822.5 – 65.8

= 756.7 kJ/kmol

38

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

39

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 Re Nu OHTC Rf ∆Tlm

E-101 1913.76 35 2.02 0.00088 9.28

E-102 695.29 30.7 1.798 0.00053 17.24

E-103 191 53.27 0.0645 0.00053 11.2

E-104 331.4 28.78 21.2 0.00053 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.

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

Plate

Heat

Exchanger

Tmilk, in = 75.7 °C Tmilk, out = 92°C

Twater, in = 100°C Twater, out = 65°C

40

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;

41

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,

𝑅𝑒 = (

957.9𝑘𝑔𝑚3 ) 9.39 𝑥 10^ − 4(0.6)

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)

ln [𝑇ℎ𝑖 − 𝑇𝑐𝑜𝑇ℎ𝑜 − 𝑇𝑐𝑖

]

42

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 1 day 1 hr 1min = 0.03699kg/s

day 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)

ln [100 − 9275.7 − 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

43

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:




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.

Plate

Heat

Exchanger

Tmilk, in = 92°C Tmilk, out =45°C

Twater, out = 72.29°C Twater, in =30°C

44

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

45

Tlm



]

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)


]

𝛥𝑇𝑙𝑚 =(92 − 72.29) − (45 − 30)

ln [92 − 72.29

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.

46

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:




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

Tubular

Heat

Exchanger

Tmilk, out = 30°C Tmilk, in = 46°C

Twater, out = 34°C Twater, in = 20°C

47

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 1 day 1hr 1min = 0.03822 kg/s

day 24hr 60min 60 sec

Thus, the Um is calculated to be

𝑈𝑚 =ṁ

𝜌𝐴𝑐

𝑈𝑚 =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

Nu = 3.66 +0.065 (

DL) Re Pr

1 + 0.04 [(DL) Re Pr]^2/3

48

Nu = 3.66 +0.065 (

0.06350.5

) 191( 7.01)

1 + 0.04 [(0.0635

0.5) (191) (7.01)]^2/3

Nu = 53.27

Tlm



]

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 =(46 − 33.5) − (30 − 20)

ln [46 − 33.530 − 20 ]

𝛥𝑇𝑙𝑚 = 11.2°𝐶

49

OHTC

𝑈 =𝑞

𝑇𝑙𝑚. 𝐴



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


E-104

Assumptions:




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.

Plate

Heat

Exchanger

Tmilk, in = 30°C Tmilk, out = 5°C

Twater, out = 17.88°C Twater, out =0.01°C

50

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 1day 1hr 1min = 0.0396 kg/s

day 24hr 60min 60s

Thus,

𝑈𝑚 =0.0396

999.8(0.04)

𝑈𝑚 = 9.90 𝑥 10^ − 4

51

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



]

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)

ln [30 − 17.88

5 − 0.01]

𝛥𝑇𝑙𝑚 = 8.03°𝐶

52

OHTC

𝑈 =𝑞

𝑇𝑙𝑚. 𝐴



To find the value of U,

𝑈 =3.4056

(8.03)(0.02)

𝑈 = 21.2 𝑘𝑊/𝑚2°C

Fouling factor


53

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 bio-

molecules. 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)

54

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)

55

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).

56

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.

57

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)

58

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).

59

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,

60

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).

61

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-3-

phosphate. This reaction is catalyzes by glyceraldehyde-3-phosphate dehydrogenase that contain

two binding sites for glyceraldehydes-3-phosphate and NAD+ to produce glycerate-1,3-

biphosphate. 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, 3-

biphosphate 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

62

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

63

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 galactose-

1-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)

64

.

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.

65

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.

66

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