integrated design

APPROVAL SHEET

This project study entitled PRODUCTION OF HIGH-VALUE ANIMAL FEEDS FROM PROTEIN-ENRICHED AGRO-INDUSTRIAL WASTES prepared and submitted by Vera Luwesa M. Allera, Mary Rose M. Estrada, Johdem I. Torayno, and Mary Jane V. Yap, in partial fulfillment of the requirements for the degree of Bachelor of Science in Chemical Engineering has been examined, accepted and passed for oral examination.

____________________________HERCULES R. CASCON, Ph. D.

MentorFaculty, ChE Department

PLANT DESIGN REPORT ON PRODUCTION OF HIGH – VALUE ANIMAL FEEDS FROM PROTEIN-ENRICHED AGRO-INDUSTRIAL WASTES

An Undergraduate Research StudyPresented to

the Faculty of the Department of Chemical EngineeringXavier University – Ateneo de Cagayan

In Partial Fulfillmentfor the Requirements for the Degree

Bachelor of Science in Chemical Engineering

byAllera, Vera Luwesa M. Estrada, Mary Rose M.

Torayno, Johdem I.Yap, Mary Jane V.

February 2015

Chemical Engineering DepartmentCollege of EngineeringXavier University – Ateneo de CagayanCorrales Avenue, Cagayan de Oro City

28 February 2015

HERCULES R. CASCON, Ph.D.SupervisorChemical Engineering DepartmentXavier University – Ateneo de Cagayan

Re: Submission of Process Design Report

Dear Sir:

With due respect and honor, we submit herewith the process design report of our study entitled “Plant Design Report on the Production of High-Value Animal Feeds from Protein-enriched Agro-industrial Wastes”. During the preparation of the report, we have tried our best to create our finest process design report and tried to show all sort of salient points of the study: the process flow diagram, material and energy balances, equipment list and unit description, profitability and environmental impact assessment.

We have collected all the updated information necessary for the improvement of our study. We hope this report will provide a clear scenario of introducing our proposed process design.

We shall be available to answer any question for clarification. Thank you for your sincere support to make our research endeavors successful.

Table of Contents

Table of Figures ……………………………….……..

……………………….................

iii

List of Tables ……………………………………………………………….................... v

1. Introduction …….………..……………………………………………...…………....1

2. Process Flow Diagram and Description………………………………….……..…...2

2.1. Process Flow Diagram Development……………………....................................3

2.1.1. Input/output Diagram……………....……….……..……………..... 3

2.1.2. Block Flow Diagram ...……………………….……..…………….... 3

2.1.3 Specifying the process …………………………………………..…. 4

2.1.4 Detailed Process Flow Diagram …………………………………….7

2.2 Process Description …………………………………………………………… 9

3. Material, Energy Balances, and Utility Requirement ……………………………. 13

3. 1 Material Balances ………...…………………………………………………… 13

3.1.1 Summary – Quantitative Flow Diagram ……..……………………. 13

3.1.2 Details of Material Balance Calculation …………………………... 15

Material Balance for Washer ………………………………………. 16

Material Balance for Air Dryer ……………………………………. 18

Material Balance for Grinding …………………………………….. 22

Material Balance for Substrate Preparation………………………... 23

Material Balance for Preparation of Starter Culture ………………. 27

Material Balance for Fermentation Vessel ………………………… 28

ii

Material Balance for Filtration …………………………………….. 33

Material Balance for Second Drying ………………………………. 35

Material Balance for Pelletizing …………………………………… 37

Accounting for unused raw material ………………………………. 38

3.2 Process Operation Scheduling ……………………………………………… 35

Calculations Involved ……………………………………………… 38

3.3 Energy Balances ……………………………………………………………. 43

3.3.1 Summary – Quantitative Flow Diagram .…………………………... 45

3.3.2 Details of Energy Balance .………………………………………… 37

Equipment .…………………………………………………………………. 46

Washer ……………………………………………………………... 46

Dryer ……………………………………………………………….. 47

Grinder ……………………………………………………………... 59

Mixer 1 …………………………………………………………….. 62

Mixer 2 …………………………………………………………….. 63

Fermentor …………………………………………………………... 63

Filter Press ………………………………………………………… 70

Vacuum Dryer ……………………………………………………… 70

Pelletizer …………………………………………………………… 71

Utilities ……………………………………………………………............... 72

Pumps ………………………………………………………………. 72

Screw Conveyor ……………………………………………………. 84

Belt Conveyor ……………………………………………………… 85

Air Heater …………………………………...………………………

85

Compressor ………………………………………………………… 86

4. Equipment Design and Specifications …………………………………………...... 91Dryer………………………………………………………………... 92

iii

Air Duct Heater……………………………………………………... 103

Mixer………………………………………………………………... 107

Fermenter…………………………………………………………… 116

5. Process Control……………………………………………………………………… 122Dryer………………………………………………………………... 122

Air Duct Heater……………………………………………………... 126

Mixer………………………………………………………………... 127

Fermenter…………………………………………………………… 130

6. Bibliography ……………………………………………………………............... 136

iv

List of Figures

Fig. 2.1 Input/output Diagram structure.…..…………………………………...... 3

Fig. 2.2 Block flow Diagram……………..……………………………………… 4

Fig. 2.3 Expanded block flow process diagram …………………….…………... 4

Fig. 2.4 More detailed block flow process diagram ………………………………... 6

Fig. 2.5 Detailed process flow diagram …………………………………………….. 8

Fig. 3.1 A quantitative flow diagram for the production of single cell protein from

agro-industrial wastes through semi-solid state fermentation using C. utilis

and S. cerevisiae ……………………………………………………………14

Fig. 3.1.2.1 Material balance around Washer ...………………….………………..……. 17

Fig. 3.1.2.2 Material balance around Dryer for Cassava Peels …………………...……. 18

Fig. 3.1.2.3 Material balance around Dryer for Pineapple Pomace ……………………. 20

Fig. 3.1.2.4 Material balance around Dryer for Ipil-ipil leaves ………………………… 21

Fig. 3.1.2.5 Material balance around Grinder for Cassava Peels ………………………. 22

Fig. 3.1.2.6 Material balance around Grinder for Pineapple Pomace ………………….. 22

Fig. 3.1.2.7 Material balance around Grinder for Ipil-ipil Leaves ……………….…….. 23

Fig. 3.1.2.8 Material balance around Mixer for Substrate Preparation …………..…….. 24

Fig. 3.1.2.9 Material balance around Starter Vessel for the Preparation of the Starter

Culture ……………………………………………………………………...

27

Fig. 3.1.2.10 Material balance around Fermentor ……………………………………... 29

Fig. 3.1.2.11 Material balance around Filter Press ………………………………………. 34

Fig. 3.1.2.12 Material balance around Pelletizer ……………...………………………. 36

Fig. 3.1.2.13 Material balance around Dryer …………………………………………... 37

v

Fig. 3.2.1 Shortened overall Ghantt chart for plant operations …………………… 40

Fig. 3.3.1 Energy balance around Washer ………...………...………...………...… 48

Fig. 3.3.2 Energy balance around Dryer ………………………………………….. 49

Fig. 3.3.3 Humidity chart for the determination of wet bulb temperature ………... 50

Fig. 3.3.4 Energy balance around Grinder ………………………………………... 62

Fig. 3.3.5 Energy balance around Fermentor ………...………...………...……….. 66

vi

List of Tables

Table 2.1. Amount of Material Entering/Generated …….……………….…….……….. 33

Table 2.2. Amount of Material Exiting (product)……………………….………………. 33

vii

Abstract

Agro-industrial wastes can be further processed to produce products with economic value

and at the same time reduce the environmental threat posed by their uncontrolled

accumulation. For this plant design, the locally available agro-industrial wastes

specifically cassava peels and pineapple pomace are utilized as raw materials for the

production of high-value animal feeds through semi-solid state fermentation. Biological

treatment was done in order to enrich the protein content of these wastes. These waste

materials are known to have low-protein content and thus, through semi-solid state

fermentation using Saccharomyces cerevisiae, their protein content is enriched. In this

study, particularly, high-value animal feeds are produced from the protein enriched agro-

industrial wastes.

The process flow of the system is established and presented in a diagram. The established

process flow diagram is incorporated with a process description in order to further

understand the process. Subsequently, material and energy balances calculations were

employed in order to determine the amount of material to be used and the amount of

product to be produced in the process. Moreover, the energy balance is employed in order

to account for the energy requirement of each process involved in the system. Both

balances can also aid the assessment of the economic feasibility of the system. Based on

literature, it is expected have a protein increase of 262% after fermentation. It was

determined that with a desired product of 10,000 kg HVAF, 23,606.09 kg of total

substrate is needed, thus, giving a yield of 42.36%.

Keywords: high-value animal feeds, agro-industrial wastes, single cell protein, semi-solid state fermentation, livestock

CHAPTER ONE

Introduction

1.1 Production of High-Value Animal Feeds from Agro-Industrial Wastes

Agro industry is an integral part of the world economic activity (Rodrigues et al.

2014) and this industry has progressed over time. Different industrial plants produce

agro-industrial products through fruit processing for various applications. The by-product

of these wastes are the so-called agro-industrial wastes which have low chemical risks,

potentially available on a large scale and can generate a biomass, rich in natural

pigments, and are potentially cost competitive (Jacob-Lopes & Franco, 2013). According

to Dhanasekaran, et al. (2011), these wastes are a renewable source of a great variety of

biotechnological potential. Currently though, due to the lack of proper handling and

utilization methods, these wastes are released directly into the environment giving a

serious impact since they emit greenhouse gases. However, such wastes are usually rich

in sugars, proteins and minerals and should be considered as raw materials for other

industrial processes. The presence of these macronutrients makes it suitable for growth of

microorganisms. One way of utilizing agro-industrial wastes is through semi-solid state

fermentation (SSSF) for the enrichment of its protein content which can be later used as

animal feed (Musatto S. I., Ballisteros, Martins, & Teixeira, 2014).

For this study, cassava peels and pineapple pomace are used as solid substrates for

the fermentation, while using ipil-ipil leaves as co-substrate for nitrogen source. The

fermentation process will be aided by the yeasts, Saccharomyces cerevisiae and Candida

utilis. Agro-industrial wastes such as cassava peels and pineapple pomace possess great

potential as raw materials for the production of different products nowadays. Animal

feeds are one of the products from further processing of these wastes. The production of

animal feed can be done through microbial fermentation. In this study, high-value animal

feeds are produced from the protein enriched agro-industrial wastes.

The product of this fermentation process is called Single Cell Protein (SCP)

which can be utilized as animal feeds. SCP refers to microbial cells grown and harvested

1

for animals or human food due to its high protein content. It may refer to the “source of

mixed protein extracted from pure and mixed culture of algae, yeast, fungi and bacteria”

(Adedayo, Ajiboye, Akintunde, & Odaibo, 2011). It can be considered as one of the

“novel” foods, those that are new to the population in question or have not been eaten in

significant amounts. Due to this, studies must be undertaken to ensure their acceptability

for human use. Studies using animals should be performed initially then followed by

closely supervised human studies. The Protein Advisory Group (PAG) of the United

Nations System has formulated guidelines to assist nutritionists and food scientists in

evaluating novel foods for human consumption. Preliminary testing includes a complete

chemical analysis of the SCP, including quantitative and qualitative information

regarding the protein, lipid, carbohydrate, vitamin, and mineral composition. Thereafter,

animal tests should be performed to determine the available energy content of the food,

quality of protein, digestibility, and availability of minerals. Experiments on animal

feeding must be performed to ensure that no adverse side effects or toxicity associated

with the use of SCP. This toxicity study include evaluation of the animal’s blood and

major organs after feeding SCP for extended periods as outlined by the PAG

recommendations (Badan Pengkajian dan Penerapan Teknologi, 1983).

Nutrients are essential components of a fermentation broth which would ensure

the growth of the microorganisms. In a fermentation process, an addition of nitrogen

source would aid the growth of the microorganisms. Leucaena leucocephala is a plant

which is commonly known as Ipil-ipil. According to the study by Escalada, R.G, it can be

considered as an efficient and cheap source of nitrogen. The leaves of ipil-ipil contain

4.3% nitrogen by weight and considerable amounts of potassium and phosphorus.

(Escalada & Ratilla, 1998) Ipil-ipil is a plant that is locally available.

1.2 Plant Location and Layout

The location of the plant is an important factor to be considered in the design of

the plant. For the production of the animal feeds, the target location of the plant is in

Manolo Fortich, Bukidnon along Sayre Highway. The site location has a coordinates of

2

8°22’14.79” North (latitude) and 124°52’11.998” East (longitude). The figure below

gives an illustration of the target site for the establishment of the plant.

Figure 1.2.1: Plant site in Manolo Fortich, Bukidnon (Google Maps, 2015).

The primary consideration considered in the selection of the location is

accessibility of the raw materials specifically cassava peels and pineapple peels and

pomace. A cassava production is located near the area enabling the easy access for

cassava peels. Bukidnon is also known due to the two major pineapple-based industries

located in the area. This allows the easy access to pineapple pomace which is a major raw

material for the animal-feed production plant. The accessibility of the raw materials will

positively contribute to the progress of the animal feed production plant. Less transport

cost can be attained due to the close distance between the processing plant and the raw

material source.

Moreover, the proposed plant site is near the Mangima River. This is illustrated in

the Figure given below. The close distance of the plant site to a body of water would

imply that there is a possible source of water for the production process.

3

Figure 1.2.2: Plant site located near Mangima River (Google Maps, 2015).

Another important consideration in the selection of the plant location is the

accessibility of forms of transport. The location is close to Sayre Highway which will

allow easy access to land transportation.

Figure 1.2.3: Plant Location close to Sayre Highway (Google Maps, 2015).

4

In terms of accessibility of the probable market, the availability of different

piggery in the area would be favorable for the animal feed production plant. The different

piggery includes backyard and commercial piggery farms.

Plant Layouts

Process units and ancillary buildings are laid out to give the most economical

flow of materials and personnel around the site. Consideration is also given to the future

expansion of the site (Seider, Seader, & Lewin, 2010). The following figure shows the

proposed site plan of thee proposed plant design.

Figure 1.2.4: Site plan.

Plant layout determines how well the plant and equipment used in the process is

laid out and is a determining factor for the economic construction and efficient operation

of a process unit (Seider, Seader, & Lewin, 2010). A proposed plant layout is shown in

the figure below.

5

Figure 1.2.5: Plant layout.

6

CHAPTER TWO

Process Flow Diagram and Description

The process flow sheet is the definitive document of the process. Thus, the

presentation needs to be clear, comprehensive, accurate and complete. This section

presents the development of the detailed process flow diagram and its process

description.

2.1 Process Flow Diagram Development

The process concept diagram or input/output diagram identifies the feed, chemical

reactions taking place, and product. For this study, cassava peels, pineapple pomace, ipil-

ipil leaves, yeast, and water are the main raw materials of the processing. These raw

materials basically undergo semi-solid state fermentation (SSSF). With this, an output of

animal feeds, carbon dioxide, water, and ethanol are generated. Fig. 2.1 below presents

the structure of the process concept diagram of the agroindustrial wastes fermentation.

Fig. 2.1: Input/output Diagram structure.

The block diagram is the simplest form of presentation. A block can represent a

single piece of equipment or a complete stage in the process. They are useful for

representing a process in a simplified form than the complicated one. (Sinnott, 2005)

From the simple input/output diagram, the block flow diagram can then be

developed. It includes the basic operations needed in the development of the desired

plant. Basically, these include feed preparation, semi-solid state fermentation,

7

purification, and pelletizing as shown in Fig. 2.2 below. Based on literature, the three

main sections of the fermentation plant to be designed are feed preparation, fermentation,

and product recovery (Simpson, Astudillo, & Acevedo, 2005) which basically is just

similar with the proposed design. The proposed detailed block flow diagram is shown in

Fig. 1. The processes involved in the main sections are basically provided in details.

Fig. 2.2: Block flow Diagram.

From the simplified block flow diagram, the processes involved are now

presented as shown in Fig. 2.3 below. Under feed preparation (labeled 1), washing,

drying, grinding, and storage of the raw materials are undertaken. Cassava peels are to be

washed due to the presence of unwanted materials such as soil and other microorganisms.

Ipil-ipil leaves and pineapple pomace are to be subjected directly to drying due to the

abundant presence of water in the latter. These raw materials are then stored separately

after the grinding process. After the feed preparation, the fermentation process is

implemented; it includes mixing of the minerals and fermentation media, preparation of

starter media, and the semi-solid state fermentation proper. Separation follows which is

just solid-liquid separation mainly, then, followed by the pelletizing process.

8

Fig. 2.3: Expanded block flow process diagram.

A more specified block glow process diagram is shown in Fig. 2.4 below. The

mixing process in square 2 is specified clearly of which another mixer is available for the

mixing of the minerals. The dissolved minerals are then added to mixer 1 where it is

mixed with the fermentation media which are just the agroindustrial wastes. A starter

vessel for the preparation of the fermentation media and the culturing of the yeast is

added. Two fermentors are basically used in this process, where one will serve as the

backup as shown in square 2 of the Fig. below. For the soli-liquid separation process in

square 3, filtration is employed, where the filter cake is of more value than the filtrate.

The pelletizing process includes pelletizing and drying of which the desired product,

animal feeds, can then be obtained.

9

1 2

4 3

Fig. 2.4: More detailed block flow process diagram.

10

1

2

3 4

From the development of the process concept diagram into a specified block flow

process diagram, a detailed process flow diagram can then be obtained. For this part, the

names of the equipment are used instead of the process itself. The labels for each

equipment are shown in Fig. 2.5 below and the description of each are presented just

above the figure. A detailed process description then follows after the detailed PFD.

Basically, as shown in the figure, all of the input materials are placed in the left side,

while the output are placed on the right side.

11

BC-101 BC-102 C-101 CR-101 CT-101 D-101 D-102 F-101 F-102 FI-103 H-101 MI-101 MX-101 MX-102 P-101Belt

Conveyor 1Belt

Conveyor 2Air

Compressor to Dryer 1

Hammer Mill for Raw Material

Size Reduction

Cooling Tower for cooling water to

Fermentor 1 and 2

Dryer for Raw

Materials

Dryer for Biomass

Fermenter 1

Fermenter 2 Filter Press for Raw

Materials

Electric Heater for Air

Supply

Pelletizer Mixer for fermentation media

and mineral solution

Mixer for liquid concentration/mineral

solution

Slurry Pump for Mineral Solution to Mixer 1

P-102 P-103 P-104 P-105 SC-101 SC-102 ST-101 ST-102 ST-103 V-101 V-102 V-103 WA-101Slurry Pump for Starter Culture to

Starter Vessel

Slurry Pump for Starter Culture to Fermentor 1 and 2

Centrifugal Pump for Cooling Water to Fermentor 1 and

2

Centrifugal Pump for Wastewater from Washer

Screw Conveyor for Ground Raw Materials

Screw Conveyor for Pelletized Biomass

Storage Tank for Cassava

Peels

Storage Tank for Pineapple

Pomace

Storage Tank for Ipil-ipil

Leaves

Starter Vessel Storage vessel for HCl Storage vessel for

NaOH

Rotary Drum Washer for

Cassava Peels

Fig. 2.5: Detailed process flow diagram.

12

2.2. Process Description

The preliminary block flow diagram for the production of high-value animal feeds is

divided into four units: Feed Preparation, Semi-solid State Fermentation, Separation, and

Pelletizing. The operations under these units are already discussed in the previous

sections. Here, the detailed process description of Fig. 2.5 is presented.

Fresh cassava peels, Stream 1, enter a polywash washer, WA-101, to remove sand and

unwanted matters. Water at room temperature and pressure of 1 atm enters WA-101.

Excess water exits WA-101 and proceeds to the wastewater treatment facility.

After washing, the cassava peels pass through a discharge conveyor belt, Stream 3, and

then enters an air dryer, D-101, where it will be heated at 55°C for 48 hours (2 days). The

drying process is also applied to other raw materials: pineapple pomace and ipil-ipil

leaves, however, it is done by batch. The cassava peels go first followed by the pineapple

pomace then ipil-ipil leaves since they have different initial moisture content and thus,

drying time. Basically, drying is done to avoid deterioration and growth of unwanted

microbes.

The dried cassava peels from the dryer then enter a grinder, CR-101, through Stream 5

with sieve (1 mm) for size reduction. Pineapple pomace and ipil-ipil leaves are also

subjected to this process. This is also done by batch. After each raw material has been

grounded, it then enters a storage tank, ST-101, by passing through a screw conveyor,

SC-101, where it will be temporarily stored prior to usage.

Mixer 1, MX-101, is where the fermentation media is prepared comprising of substrates

(cassava peels and pineapple pomace) and co-substrate (ipil-ipil leaves) coming from

Stream 7 and mineral solution coming from Stream 8. The moisture content (liquid

concentration) is prepared in Mixer 2, MX-102, and enters through Stream 8 into MX-

101. It is where distilled water and other minerals (Stream 15) are mixed to produce a

mineral solution. The minerals are composed of (NH4)2SO4, MgSO4∙H2O, KH2PO4, and

MnSO4 which are fed manually due to their negligible amounts.

9

A simultaneous process occurring in one area of the plant is the microorganism and

inoculum preparation. The microorganism S. cerevisiae, Stream 11 are fed to the starter

vessel, V-101, where a portion of the fermentation media will be fed through Stream 10.

After which, the prepared starter culture is then fed to Fermentor 1 and 2 through Stream

13.

Prior to use, Fermenter 1 and 2, are injected with steam for sterilization at 121 °C for 20

minutes. The fermentation media then enters the fermentor. The inoculum prepared from

starter vessel, V-101, enters F-101/2, where it is then inoculated to the fermentation

media where it will be fermented for 3 days at 30°C to carry on the semi-solid state

fermentation.

After the 3-day fermentation, the slurry then enters a filter press, FI-101, to force separate

the water from the slurry at room temperature, 25°C. The water exits the unit and the

single cell protein enters a pellet mill, MI-101. The pellet mill presses the SCP and gives

the pellet (cylindrical) shape of the product under temperature, pressure, and moisture. It

leaves the mill through Stream 17 and passes a screw conveyor for storage.

14

CHAPTER THREE

Material, Energy Balances, and Utility Requirement

This chapter presents the material balance, energy balance, and utility requirement

calculations for the proposed plant design. It is important to have detailed quantitative

calculations of the materials required since it would eventually determine the feasibility

of the design.

3.1. Material Balances

One important matter to be considered in producing products is the material

balance. It involves the accounting of the input (raw materials) and the output (products,

byproducts, wastes, etc.) of the processes. This section provides a material balance of the

processes involved which will be provided in the succeeding sections of this paper. It will

provide an overview of the accounting of the amount of raw materials (agro-industrial

waste, water, minerals and yeast) that will be used and the corresponding amount of high-

value feeds that will be produced in each process. Moreover, the amount of by-products

that will be produced in the process will also be calculated. The plant design starts from

the preparation of raw materials which includes washing, drying, grinding, and storage.

Then, fermentation procedure is employed in order to increase the protein content of the

substrate. Separation is then employed in order to separate the products and by-products.

The process then continues to the product separation and by-product purification.

In solving for the material balance of the system and checking if the objectives are

met, the protein content increase of the substrate is to be calculated. Available values of

moisture content are used to determine the value as to which the raw materials are dried.

Ethanol and CO2 yield of the substrates are based on literature. To determine certain

values which have no existing literature to validate, calculations were carried out. The

amount of protein produced with respect to the substrate is basically the main aim of this

fermentation process.

15

3.1.1 Summary – Quantitative Flow Diagram

A quantitative summary of the materials that enter and leave the processes are

shown in Fig. 3.1 below. A sum of 61,968.14 kg input and output is obtained. The

breakdown of the amount of these materials is shown in the Fig. below. This is then

followed by a detailed calculation of the values.

Basis: 1 operating batch

Unit designed to produce 10,000 kg HVAF per day

cassava peels 5,684.35 5,115.91 waterwater 5,684.35 6,252.78 kg CP

pineapple pomace 7,431.21 2,163.03 water from CPipil-ipil leaves 1,240.63 4,089.76 kg CP 3,341.45 water from PP

4,089.76 kg PP 72.13 water from IL1,168.50 kg IL

4,089.76 kg CP4,089.76 kg PP1,168.50 kg IL

23,370.03 water (for substrate) 12,853.52 (NH4)2SO4 35.06 23,606.09 kgMgSO4 10.52 MnSO4 1.17 542.94 ethanolKH2PO4 18.70 22,520.21 kg 542.94 CO2Water (for MS) 1,103.07 Yeast culture 236.06 8,234.50 recovered liquid

14,285.71 kg

3,571.43 water10,000.00 kg 714.29 ethanol

10,000.00 HVAF

SUM= 34,298.61 kg SUM= 34,298.61 kg

OUTPUT (kg)INPUT (kg) PROCESS

Filtration

Drying

Pelletizing

Washing

Drying

Grinding

Mixing

Fermentation

Fig. 3.1.: A quantitative flow diagram for the production of high-value animal feeds from

protein-enriched agro-industrial wastes.

16

3.1.2 Details of Material Balance Calculation

Material balance is employed in order account for all material that enters and

leaves each process. For this process, high value animal feeds are the product to be

produced. It is produced from fermented agro-industrial wastes namely cassava peels and

pineapple pomace. Initially, an overall material balance for the entire system is done.

Figure 3.1 gives an illustration of the general input and output streams of the system. The

basis for the entire calculation is 10,000 kg high-value animal feeds (HVAF). Production

of the animal feeds is not limited by the availability of the raw materials. The two main

raw materials, namely, cassava peels and pineapple pomace, are readily available due to

the use of cassava and pineapple as raw materials in different industries. In the

succeeding discussion below, a material balance is employed for each major process

involved.

BASIS: 10,000 kg HVAF/day

a. Pelletizer

A product preparation step which is pelletizing is employed. It is to form the final

product into pellets to be dried which will then be ready for consumption. For the

material balance calculation of the pelletizing process, the following assumptions are

considered:

1. Negligible mass loss.

2. Steam does not affect the mass of the product.

Figure 3.2.1: Material balance around Pelletizer.

Calculation:

m¿=mout

17

mmeal=?

mmeal=10,000 kg meal/day

b. Dryer 2

Drying is employed to further reduce the moisture content of the filtered material

and to eliminate the remaining ethanol content of the same material. For the process, air

is the heat medium. It is desired to have a moisture content of 18% after drying. The

following assumptions are taken into consideration:

1. The output material has 18% moisture content.

2. No ethanol remains in the liquid component of the fermented after drying.

3. 5% ethanol is removed.

4. 25% water is removed from the meal.

Figure 3.1.2.2: Material balance around Dryer 2.

Calculating for amount of filtered meal (mfm) fed to dryer,

mfm=mmeal+mlr+methanol (Eqn. 1)

where

mlr=0.25mfm (Eqn. 2)

methanol=0.05 mfm (Eqn. 3)

Substitute Eqns. 2 and 3 to 1,mfm=mmeal+mfm (0.25 )+mfm (0.05 )

mfm=10,000+mfm (0.25 )+mfm (0.05 )

18

mfm (1−0.25−0.05 )=10,000 kgmfm=14,285.71 kg

Liquid removed (Eqn. 2),

mlr=0.25 (14,285.71kg)

mlr=3,571.43 kg

Ethanol removed (Eqn. 3),

methanol=0.05(14,285.71 kg)

methanol=714.29 kg

c. Filtration

Filtration is a process to employ solid – liquid separation. There are different

equipments that can be used to employ filtration. For the given system, the filtration

process is employed through a filter press. The main objective of this process is to

reduce the liquid component of the fermented material. The following assumptions

are considered in the calculation:

1. 70% of the liquid component is removed.

2. 10% of fermented material is lost. The material could either be in the filter press

or in the filtrate.

3. The equipment used for filtration is filter press.

4. The material after filtration contains around 25 to 70% solid content (National

Metal Finishing Resource Center).

5. Approximately 47% of the fermented substrate is liquid (based on previous

calculations).

6. 65% of the liquid from the fermented substrate is removed or recovered (filtrate).

35% goes with the filtered meal.

19

Figure 3.1.2.3: Material balance around Filter Press.

Calculations:

OMB:

m¿=mout

mfs=mlr+mlm

mfs=mlr+14,285.71 (Eqn. 4)

(where fs=fermented substrate, lr=liquid recovered, fm=filtered meal)

Liquid balance:

mfs(0.47)=mlr+14,285.72 kg (0.35 )(0.47) (Eqn. 5)

Solving equations 1 and 2 simultaneously gives,

mfs=22,520.21 kg

mlr=8,234.50 kg

Liquid unrecovered

mlm=(0.35 )( .47)(14,285.71kg )

mliquid∈the meal=2,350.00 kg

Solids

msolids=14,285.71 kg−2,350.00 kg

msolids=11,935.71kg

Percent solid composition after filtration

% solids=msolid component out

mfiltrationoutput

×100

20

% solids= 11,935.71kg11,935.71kg+8,234.50 kg

× 100

% solids=60 %

which falls within the range set by (National Metal Finishing Resource Center).

d. Fermenter

The most important process in the production of high-value animal feeds is the

fermentation. The fermentation is employed in order to increase the protein content of the

substrate through the use of yeast. The enrichment is done through semi-solid state

fermentation. In this process, the substrate contains 55% water. Two fermentation setups

are to be employed for the process. Both setups will have the same amount of substrate

and starter culture. Different factors are considered for the process and the following

assumptions are taken into consideration:

1. Initial protein content of raw materials: cassava peels is 4.21%, pineapple pomace

is 6.4% and ipil-ipil leaves is 23%.

2. There is a 262% increase in the protein content of the material (Gelinas &

Barrette, 2007).

3. 424 mmol CO2 is produced per 100 g of cell biomass produced (Jeppson, Yu, &

Hahn-Hagerdal, 1996).

4. Ratio of ethanol to carbon dioxide is 1 mol:1mol (Jeppson, Yu, & Hahn-Hagerdal,

1996).

5. The substrate will be equally distributed in each fermentor.

6. 50% of the yeast is protein.

7. The amount of biomass produced is equal to the amount substrate consumed.

8. Substrate consumed = biomass added.

9. 2.3% of ethanol and CO2 is generated from the total substrate.

21

Figure 3.1.2.4: Material balance around Fermenter.

OMB:

msubstrate=mC O2+methanol+22,520.21 kg (Eqn. 6)

where

mC O2=0.023 m substrate (Eqn. 7)

methanol=0.023 msubstrate (Eqn. 8)

Substitute Eqns. 7 & 8 to 6,

msubstrate=0.023 msubstrate+0.023 msubstrate+22,520.21 kg

msubstrate=23,606.09 kg

For Protein

Before fermentation:

moriginal protein=(mCP dried) (0.0421 )+( mPP dried ) (0.0640 )+(mIL dried )(0.2300)(Eqn. 9)

After fermentation:

maccumulated protein=0.5 m yeast +3.62 moriginal protein (Eqn. 10)

For fermented meal

mfermented meal=mtotal substrate−msubstrate consumed−mtotal water added+mbiomass added(Eqn. 11)

mC O2=22,520.21 kg ( 0.023 ) msubstrate=542.94 kg=methanol

22

Biomass added

mbiomass added=maccumulated protein

0.50

(Eqn. 12)

For starter culture

mstarter culture=0.01msubstrate

mstarter culture=0.02(23,606.09 kg)

mstarter culture=471.12 kg

where half of the culture is yeast.

Original mass of substrate before culture is added

msubstr ate=23606.09 kg−472.12 kg2

msubstrate=23,370.03 kg

Moreover, the mass of the total biomass is calculated:

mtotal biomass added=mtotal proteinout ×1 biomass0.5 protein

Eq. 13

Then, the percent ethanol in the liquid component can then be calculated:

% ethanol∈liquid component=methanol produced

mliquid component

×100Eq. 14

e. Starter Culture Tank

A starter culture is prepared in order to allow the yeast to adjust to the

fermentation conditions that to be adopted. The same substrate that will be used in the

fermentation procedure is introduced in the starter vessel.

Assumptions:

23

1. 1% of the substrate from the mixer will be used for the inoculation.

2. Mass of yeast added is 1% of the total substrate. Therefore, there is 1:1 mass

proportion.

3. Yeast culture is added not pure yeast.

Figure 3.1.2.5: Material balance around Starter Culture Tank.

Mass of substrate to culture,

msubstrate ¿culture ¿=msubstrate (0.10 ) (Eqn. 15)

Mass of yeast for substrate, m yeast=msubstrate (0.10 ) (Eqn. 16)

Calculating mass of starter culture, msc=msubstrate ¿culture ¿+m yeast=msubstrate (0.10 )+msubstrate (0.10 ) (Eqn. 17)

(where sc=starter culture)

Thus,

m yeast=0.01 (23,606.09 kg )

m yeast=236.06 kg=msubstrate ¿culture ¿

msc=236.06 kg+236.06 kg

msc=472.12 kg

f. Mixing 1

Then, mixing of the components of the substrate is done. Minerals, solid agro-

industrial wastes and water is added in the mixers. The mixer for minerals allows the

mixing of minerals prior to introduction to the total mixer. Four minerals are to be added

24

and diluted with water. The total mixer on the other hand, allows the mixing of all

components of the substrate namely, ground raw materials, water and mineral solution.

The following assumptions are considered in the calculation:

1. The total substrate is composed of 17.5 % cassava peels, 17.5% pineapple

pomace, 5% ipil-ipil leaves, 5% minerals and 55% water.

2. 94.4% of mineral solution is water.

3. Mineral solution: 1.5 g (NH4)2SO4/kg substrate, 0.05 g MnSO4/ kg substrate, 0.8 g

KH2PO4/kg substrate and 0.45 g MgSO4/kg substrate.

Figure 3.1.2.6: Material balance around Mixer 1.

Assumptions:

msubstrate=mCP+mPP+mIL+mH 2 O+mMS

23,370.03 kg=mCP+mPP+mIL+mH 2 O+mMS (Eqn. 18)

where

mCP=0.175 (msubstrate ) (Eqn. 19)

mPP=0.175 (msubstrate ) (Eqn. 20)

mIL=0.05 ( msubstrate ) (Eqn. 21)

mH2 O=0.55 (msubstrate ) (Eqn. 22)

mMS=0.05 (msubstrate ) (Eqn. 23)

25

where mMS,

mMS=m( N H 4)2 SO 4+mMnS O4

+mK H 2 PO4+mMgS O4

m( N H 4)2 S O4=

1.5 g ( N H 4 )2 S O4

kg substrate

(Eqn. 24)

mMnSO 4=

0.05 g MnS O4

kgsubstrate

(Eqn. 25)

mK H 2 P O4=

0.8 g K H 2 PO 4

kgsubstrate

(Eqn. 26)

mMgSO 4=

0.45 g MgS O4

kgsubstrate

(Eqn. 27)

mH2 O added=0.944 kg H 2 O

kg MS

(Eqn. 28)

Total water addedmtotal water=mH 2 Oadded ¿ MS ¿+mH 2 O¿ substrate ¿

mtotal water=0.994 kg H 2O

kg MS+0.55 (msubstrate )

(Eqn. 29)

Thus, substitute calculated values to Eqns. 19-29,

mCP=0.175 (23,370.03 kg )=4,089.76 kg

mPP=0.175 (23,370.03 kg )=4,089.76 kg

mIL=0.05 (23,370.03 kg )=1,168.50 kg

mH2 O=0.55 (23,370.03 kg )=12,853.52 kg

mMS=0.05 (23,370.03 kg )=1,168.50 kg

m( N H 4)2 S O4=

1.5 g ( N H 4 )2 S O4

kg substrate(23,370.03 kg )( 1 kg

1000 g )=35.06 kg ( N H 4 )2 SO4

mMnSO 4=

0.05 g MnS O4

kgsubstrate(23,370.03 kg )( 1 kg

1000 g )=1.17 kg MnSO 4

26

mK H 2 P O4=

0.8 g K H 2 PO 4

kgsubstrate(23,370.03 kg )( 1kg

1000 g )=18.70 kg K H 2 P O4

mMgSO 4=

0.45 g MgS O4

kgsubstrate(41,928.73 kg )( 1 kg

1000 g )=18.87 kg MgS O4

mH2 O added=0.944 kg H 2 O

total MS(1.168 .50 kg )=1,103.07 kg H 2O

Now, the mass of original protein from the substrate can be determined since the

amounts of raw materials to be used are obtained.

Before fermentation (Eqn. 9):

moriginal protein=(4,089.76 kg ) (0.0421 )+(4,089.76 kg ) (0.0640 )+(1,168.50 kg )(0.2300)

moriginal protein=702.68 kg

After fermentation (Eqn. 10):

maccumulated protein=0.5 m yeast +moriginal protein

maccumulated protein=moriginal protein (2.62 )−m yeast (0.50 )

maccumulated protein=702.68 kg (2.62 )−236.06 kg (0.50)maccumulated protein=1,722.99 kg

g. Grinding

Grinding is employed in order to reduce the particle size of the raw materials. Smaller

size for raw materials would allow the microorganisms during fermentation to easily

access the nutrients available in the raw materials. The following assumptions are

considered in the calculation:

1. No material is lost during the process. Thus, input is equal to the output.

2. Each material is ground separately.

Cassava Peels

27

Figure 3.1.2.7: Material balance around Grinder for Cassava Peels.

mdried CP=mground CP (Eqn. 30)mdried CP=4,089.76 kg

Pineapple pomace

Figure 3.1.2.8: Material balance around Grinder for Pineapple Pomace.

mdried PP=mground PP (Eqn. 30)mdried PP=4,089.76 kg

Ipil-ipil Leaves

Figure 3.1.2.9: Material balance around Grinder for Ipil-ipil Leaves.

mdried IL=m ground IL (Eqn. 31)

mdried IL=1,168.50 kg

h. Drying

Drying is employed after washing in order to achieve desired moisture content of

the raw material for the succeeding processes. For this system, the heating medium is

dry air. The air is heated by a fired heater. The drying process is employed at 55 °C. It is

28

important to consider the temperature of the system to avoid degradation of important

components of the raw material that is significant to the succeeding processed. Moreover,

the drying process is employed in batch condition. Each raw material is dried separately.

The following assumptions are considered for the calculation:

1. Each raw material is dried separately

2. After drying, 14% moisture content remains

3. Original moisture content (wet basis) of raw materials is 27% for cassava peels,

35.4% for pineapple pomace and 16% for ipil-ipil leaves.

Cassava Peels

Figure 3.1.2.10: Material balance around Dryer 1 for CP.

Initially, based on literature, 27% moisture content is available in the cassava

peels. Calculate moisture content dry-basis,

% moisturecontent drybasis= 0.271−0.27

×100

% moisturecontent drybasis=37 %

Accounting the moisture absorbed by the peels during washing,

% initial moisture content=43.75 %

Solids balance:

29

mwashed CP (1−0.4375 )=mdriedCP (1−0.14 )

mwashed CP (1−0.4375 )=4,089.76 kg (1−0.14 )

mwashed CP=6,252.78 kgCP

Water removed

mwater=6,252.78 kg−4,089.76 kg=2,163.03 kg H 2O

Pineapple pomace

Figure 3.1.2.11: Material balance around Dryer 1 for PP.

Initially, moisture content is 34.5% dry basis, calculating for the percent moisture

content in dry basis of the pineapple pomace,


×100

% moisturecontent drybasis=52.67 %

Calculate mass of pineapple pomace to be dried,

mPP (1−0.5267 )=mdried PP(1−0.14)

mPP (1−0.5267 )=4,089.76 kg (1−0.14)

mPP=7,431.21 kg PP

Water removed

mwater=7,431.21 kg−4,089.76 kg=3,341.45 kg H 2 O

30

Ipil-ipil Leaves

Figure 3.1.2.12: Material balance around Dryer 1 for IL.

Calculating for the percent moisture content in dry basis of the ipil-ipil leaves,


×100

% moisturecontent drybasis=19 %

Solids balance:mIL (1−0.19 )=mdried IL(1−0.14)

mIL (1−0.19 )=1,168.50 kg(1−0.14)

mIL=1,240.63 kg IL

Water removed:

mwater=1,240.63 kg−1,168.50 kg=72.13 kg H 2O

i. Washing

Washing is employed in order to remove unnecessary solid particles (i.e. soil)

from the raw materials. For this process, only cassava peels is washed. Moreover, the

following assumptions are considered in the calculation:

1. Meyers’ PolywashTM Multi-Produce Washers is to be used for washing.

31

2. 10% of the water used remains in the material.

3. Only cassava peels will be washed.

4. Ratio of water and cassava peels is 1:1. This is to ensure that there is

enough water added to remove the impurities.

5. The amount of waste removed is negligible.

6. 10% of the moisture (water) is absorbed by the cassava peels.

Figure 3.1.2.13: Material balance around Washer.

Calculating mass of cassava peels fed to washer,

mCP=mH 2 O

mwashed CP=mCP (1.10) (Eqn. 32)

mCP=6252.78 kgCP

1.10

mCP=5,684.35 kg=mH 2 O

Water out

mH2 O out=0.90(5,684.35 kg )

32

mH2 O out=5,115.91kg3.3 Energy Balances

Energy balance is one of the important requirements for the design of an

industrial plant. For this study, it is important to calculate the heat and power

requirements of the equipment and utilities involved in the processing of the agro

industrial wastes into high-value animal feeds. This is done to validate if the plant is

feasible and has economical value for materialization in the future. An energy balance of

the heat and power requirements for the processing of the agro-industrial wastes into

high-value animal feeds is done in this report. Also, calculations for the power

requirement for washing, belt and screw conveyors, milling, and pelletizing are

presented. These will be shown in the succeeding sections of this paper.

In this energy balance, the determination of the energy requirements of the

processes is to be specified and accounted using available data from journal articles,

Perry’s Chemical Engineering Handbook, from calculations described in books by

Timmerhaus, Seider, and other available online sources.

The assumptions made were the following:

1. The water used for washing is not absorbed by the raw materials. Thus, mwater

in=mwater out.

2. There are no mass losses in the grinding of the raw materials.

3. Only CO2 leaves the fermenter during fermentation.

3.3.1 Summary – Quantitative Flow Diagram of Power/Heat Requirement

A summary of the quantitative flow of the energy and power requirements is

shown in the figure below.

33

Table 3.3.1: Summary of power requirements for the various equipments and utilities and

the amount of utilities to be used per batch of high value animal feeds produced.

Equipments and Utilities Number of

Units

Power/ Heat

Requirement

Amount (kg)

A. Electric Power

Air Heater 1 275.98 kW -

Belt conveyor 2 1.94 kW -

Belt conveyor (Washer) 1 0.916 kW -

Blower 1 10.39 kW -

Compressor 1 202.51 kW -

Filter Press 1 90 kW -

Grinder 1 36 kW -

Mixer 1 1 76.43 kW -

Mixer 2 1 0.023 kW -

Pelletizer 1 325 kW -

Pump (Cooling Water) 1 0.35 kW -

Pump (Washer Feed) 1 0.122 kW -

Pump (Drain Water in Washer) 1 0.54 kW -

Pumps (Slurry) 8 128 kW -

Screw Conveyor 1 2.23 kW -

Tunnel Dryer 1 11.4 kW -

B. Cooling Water (Tin = 20°C,

ΔT = 10°C)

- - 240,813.87

3.3.2 Details of Energy Balance Calculation

The energy balance of the processes is presented in the succeeding sections.

i. Equipment

A. Washer

Fresh cassava peels are first washed to remove unwanted particles such as soil.

Assumptions:

1. It is assumed that a PolywashTM Washer is employed for washing.

34

Cassava Peels

Figure 3.3.2.1: Energy balance around Washer.

The Polywash™ uses advanced technology for washing and cleaning of fruit,

vegetables, leafy products, tubers, roots, seafood, fish, and more, hence, it is applicable

for cassava peels.

Figure 3.3.2.2. Mechanism of a Polywash™ Washer.

The Polywash™ uses an innovative combination of turbulent and calm zones to

produce a highly effective wash while removing fine debris that would otherwise be

35

carried out with the product. Sediment is flushed from the system periodically through

the main waste valve.

The Polywash™ has a capacity of up to 50 tons/hr or 13.89 kg/s.

In this equipment, it does not give a specific power requirement; however, the

power required will be used for the pump, blower and discharge belt conveyor. The

discussion on the power requirement will be shown the utilities section.

B. Dryer

Drying of the feed is basically done to increase the percent dry matter content

which has little effect on the crude protein.

Figure 3.3.2.3: Energy balance around Dryer.

In the calculation for the heat balance in the dryer, it is initially assumed that the

air and the raw material have a counter current flow. The conditions of the raw material

are known. The air is the heating medium for the system. Since the conditions for the dry

air are not known then the following assumptions for the dry air are considered:

T A1=120 °C

H 1=0.05kgH2 O

kgDA

NTU = 1.75

Then, the temperature of the outlet of the air stream, TA2, can then be calculated.

The equation used in determining the outlet temperature of the air is given as:

NTU=ln(T A2−tw

T A1−tw) Eq. 64

36

wheretw is the wet bulb temperature of the air. The value for the wet bulb temperature of

air can be taken from the humidity chart for air and gas given in the Principles of

Transport Processes and Separation Processes by Geankoplis, 2003.

Figure 3.3.2.4: Humidity chart for the determination of wet bulb temperature.

Given that the inlet temperature of air is 120°C and the humidity is 0.05 kg water vapor/

kg dry air, it can determined using the humidity chart that the wet bulb temperature of the

system is 48°C.

Calculating for the outlet temperature of the air:

NTU=ln(T A2−tw

T A1−tw) Eq. *

2.0=ln( 120−48T A2−48 )

T A2=57.7441O C

Three different materials will be dried in the process: cassava peels, pineapple

pomace and ipil-ipil leaves. Each material requires different amount of heat for drying.

First, the calculation of the heat balance for drying the cassava peels was employed.

37

Cassava peels

Considering the equation for the material balance on the moisture of the air and

the cassava peels:

mA1 H 1+ms 2 x2=mA2 H 2+ms 1 x1 Eq. 65

From the overall material balance within the dryer, the equation is given by:

mA1+ms 2=mA 2+ms 1 Eq. 66

The following variables are known:

ms 2=4,089.76 kgcassava peels

x2=0.14 kg totalmoisture /kgdry solid

H 1=0.05kg watervapor

kgdry air

ms 1=6,252.78 kgcassava peels

x1=0.4375 kg total moisture /kg dry solid

Substituting known values to the equations gives,

mA1 H 1+4,089.76 (0.14 )=mA2(0.05)+6,252.78 (0.4375) Eq. 67

And

mA1+4,089.76=mA 2+6,252.78 Eq. 68

The, the enthalpy of the gas in the outlet stream in kJ/ kg dry air was then

calculated using the equation

HG2' =cs ( T A 2−T o )+ H 2 λo

Eq. 69

where cs is the humid heat given by the expression cs=1.005+1.88 H , λo is the latent

heat of water at temperature To. The To of the system is 0°C which corresponds to a value

of λo equal to 2501 kJ/kg. Substituting known values:

HG2' = [1.005+1.88 (.050 ) ] (120−0 )+(2501∗0.050 )

HG2' =256.93

kJkgDA

Then, the enthalpy of the gas in the inlet stream was calculated.

HG1' =cs (T A1−To )+ H 1 λo

where TA1 is 57.7441°C, the same expression for cs and the same value for To and λo.

38

HG1' = [1.005+1.88 (.050 ) ] (57.7441−0 )+(2501 H 1 )

HG1' =63.4607659+2501 H 1

In calculating for the enthalpy of the wet solid raw material in the inlet, the

equation used was:

H s 1' =c ps (T S 1−T o )+x1 cPA(T s 1−T o) Eq. 70

where cps is the specific heat of cassava peel , 2.455 kJ/kg °C, TS1 is equal to 25°C, To is

equal to 0°C, x1 is equal to 0.4375 and cPA is the specific heat of the water in the raw

material, 4.187 kJ/kg. Substituting known values:

H s 1' =2.455 (25−0 )+(0.4375)(4.187)(25−0)

H s 1' =107.1703

kJkg

For the enthalpy of the wet solid in the outlet stream, same calculations were

applied, only that the value for TS2 is 60°C and the fraction of water in the material, x2 is

0.14.

H s 2' =c ps (T S 2−T o )+x2 cPA(T s 2−T o)

H s 2' =2.455 (60−0 )+(0.14)(4.187)(60−0)

H s 2' =182.4708

kJkg

Then, the calculation for the amount of dry air to be used in the system can be

employed. Given the equation:

mA2 HG2' +ms 1 H s 1

' =mA1 HG 1' +ms 2 H s 2

' Eq. 71

Substituting known values:

mA2 (256.93 )+(6,252.78 ) (107.1703 )=mA 1 (63.4607659+2501 H 1 )+( 4,089.76 ) (182.4708 )Eq. 72

Thus, using equation 7:

mA1+4,089.76=mA 2+6,252.78

mA1=mA 2+2163.02 Eq. 73

Substituting this to equations 6 and 11 gives:

mA1 H 1+4,089.76 (0.14 )=mA2 (0.05 )+6,252.78 (0.4375 )

39

(m¿¿ A 2+2163.02)H 1+4,089.76 (0.14 )=mA 2(0.05)+6,252.78(0.4375)¿

H 1=0.05mA2+2163.02

mA2+2163.02

Eq. 74

Substituting the equations 12 and 13 to equation 11 gives:

mA2 (256.93 )+(6,252.78 ) (107.1703 )=mA 1 (63.4607659+2501 H 1 )+( 4,089.76 ) (182.4708 )

mA2 (256.93 )=mA 1 (63.4607659+2501 H 1 )+76149.47

mA2 (256.93 )=(mA2+2163.02)(63.4607659+2501( 0.05 mA2+2163.02mA2+2163.02 ))+76149.47

mA2=82,186.38 kgdry air

Substituting the value to equations 12 and 13 gives:

mA1=mA 2+2163.02

mA1=82,186.38+2163.02

mA1=84,349.40 kg DA

and

H 1=0.05mA2+2163.02

mA2+2163.02

H 1=0.05(82,186.38)+2163.02

(82,186.38)+2163.02

H 1=0.074kgwatervapor

kgdryair

Pineapple Pomace


the pineapple pomace:

mA1 H 1+ms 2 x2=mA2 H 2+ms 1 x1 Eq. 75




ms 2=4,089.76 kg pineapple pomace

40


H 1=0.05kgwat ervapor

kgdryair

ms 1=7,431.21 kg pineapple pomace

x1=0.5267 kg totalmoisture /kg dry solid


mA1 H 1+4,089.76 (0.14 )=mA2(0.05)+7,431.21(0.5267) Eq. 77

And

mA1+4,089.76=mA 2+7,431.21 Eq. 78



HG2' =cs ( T A 2−T o )+ H 2 λo

Eq. 79

where cs is the humid heat given by the expression cs=1.005+1.88 H , λo is the latent



HG2' = [1.005+1.88 (.050 ) ] (120−0 )+(2501∗0.050 )

HG2' =256.93

kJkgDA




HG1' = [1.005+1.88 (.050 ) ] (57.7441−0 )+(2501 H 1 )

HG1' =63.4607659+2501 H 1


equation used was:


where cps is the specific heat of pineapple pomace and assumed that it is the same with

cassava peels since they are both biomass, 2.455 kJ/kg °C, TS1 is equal to 25°C, To is equal

to 0°C, x1 is equal to 0.5267and cPA is the specific heat of the water in the raw material,

4.187 kJ/kg. Substituting known values:

41

H s 1' =2.455 (25−0 )+(0.5267)(4.187)(25−0)

H s 1' =116.5073

kJkg



0.14.


H s 2' =2.455 (60−0 )+(0.14)(4.187)(60−0)

H s 2' =182.4708

kJkg




' =mA1 HG 1' +ms 2 H s 2

' Eq. 81


mA2 (256.93 )+(7,431.21 ) (116.5073)=mA 1 (63.4607659+2501 H 1 )+( 4,089.76 ) (182.4708 )Eq. 82


mA1+4,089.76=mA 2+7,431.21

mA1=mA 2+3341.45 Eq. 83


mA1 H 1+4,089.76 (0.14 )=mA2 (0.05 )+7,431.21 (0.5267 )

(mA2+3341.45) H 1+4,089.76 (0.14 )=mA2 (0.05 )+7,431.21 (0.5267 )

H 1=0.05mA2+3341.45

mA2+3341.45

Eq. 84


mA2 (256.93 )+(7,431.21 ) (116.5073)=mA 1 (63.4607659+2501 H 1 )+( 4,089.76 ) (182.4708 )

mA2 (256.93 )=mA 1 (63.4607659+2501 H 1 )−119,528.43

42

mA2 (256.93 )=(mA2+3341.45)(63.4607659+2501( 0.05 mA2+3341.45mA2+3341.45 ))−119,528.43

mA2=123,495.81 kg dry air


mA1=mA 2+3341.45

mA1=123,495.81+3341.45

mA1=126,837.26 kg DA

and

H 1=0.05mA2+3341.45

mA2+3341.45

H 1=0.05(123,495.81)+3341.45

123,495.81+3341.45


kgdryair

Ipil – ipil Leaves


the ipil – ipil leaves:

mA1 H 1+ms 2 x2=mA2 H 2+ms 1 x1 Eq. 85




ms 2=1,168.50 kg ipil−ipil leaves



kgdryair

ms 1=1,240.63 kg ipil−ipil leaves

x1=0.19 kg total moisture /kgdry solid


mA1 H 1+1,168.50 (0.14 )=mA 2(0.05)+1,240.63(0.19) Eq. 87

And

43

mA1+1,168.50=mA 2+1,240.63 Eq. 88



HG2' =cs ( T A 2−T o )+ H 2 λo

Eq. 89

where cs is the humid heat given by the expression cs=1.005+1.88 H , λO is the latent



HG2' = [1.005+1.88 (.050 ) ] (120−0 )+(2501∗0.050 )

HG2' =256.93

kJkgDA




HG1' = [1.005+1.88 (.050 ) ] (57.7441−0 )+(2501 H 1 )

HG1' =63.4607659+2501 H 1


equation used was:


where cps is the specific heat of ipil – ipil leaves and has the same value with the cassava

peels, 2.455 kJ/kg °C, TS1 is equal to 25°C, To is equal to 0°C, x1 is equal to 0.5267and cPA

is the specific heat of the water in the raw material, 4.187 kJ/kg. Substituting known

values:

H s 1' =2.455 (25−0 )+(0.19)(4.187)(25−0)

H s 1' =81.26325

kJkg



0.14.


H s 2' =2.455 (60−0 )+(0.14)(4.187)(60−0)

44

H s 2' =182.4708

kJkg




' =mA1 HG 1' +ms 2 H s 2

' Eq. 91


mA2 (256.93 )+(1240.63 ) (81.26325 )=mA 1 (63.4607659+2501 H 1 )+(1168.50 ) (182.4708 )Eq. 92


mA1+1168.50=mA2+1240.63

mA1=mA 2+72.13 Eq. 93


mA1 H 1+1168.50 (0.14 )=mA2(0.05)+1240.63(0.19)

(mA2+72.13)H 1+1168.50 (0.14 )=mA2(0.05)+1240.63(0.19)

H 1=0.05mA2+72.13

mA2+72.13

Eq. 94


mA2 (256.93 )+(1240.63 ) (81.26325 )=mA 1 (63.4607659+2501 H 1 )+(1168.50 ) (182.4708 )

mA2 (256.93 )=mA 1 (63.4607659+2501 H 1 )+112,399.50

mA2 (256.93 )=( mA 2+72.13 )(63.4607659+2501 (0.05 mA 2+72.13mA 2+72.13 ))+112,399.50

mA2=4,346.35 kgdry air


mA1=mA 2+72.13

mA1=4,346.35+72.13

mA1=4,418.48 kg DA

45

and

H 1=0.05mA2+72.13

mA2+72.13

H 1=0.05(4,346.35)+72.13

4,346.35+72.13


kgdryair

Table 3.3.2: Summary of the Humidity, H1, leaving and mass of dry air, mA1, required in

each drying process.

Raw Material Humidity of air exiting, H1

(kg water vapor/kg dry air)

Mass of Dry Air

(kg)

Cassava Peels 0.074 82,186.38

Pineapple Pomace 0.075 123,495.81

Ipil-ipil Leaves 0.0655 4,346.35

Total 210,028.54

In calculating for the heat required for the drier, the equation below is used given

that the heat capacity of air at 120°C is 1.013 kJ/kg·K ( (Air Properties, 2014):

H=mA cp , A ∆ T

In drying the cassava peels for 18.783 hrs, the amount of heat required is:

H=(82,186.38 kg )(1.013

kJkg·K ) (120−57.7441 ) K

18.783 hrs (3600 s1 hr )

H=76.65 kW

In drying the pineapple pomace for 28.224 hrs, the amount of heat required is:

H=(123,495.81 kg )(1.013

kJkg·K ) (120−57.7441 ) K

28.224 hrs (3600 s1 hr )

H=76.65 kW

In drying the ipil – ipil leaves for 0.933 hrs, the amount of heat required is:

46

H=( 4,346.35 kg )(1.013

kJkg·K )(120−57.7441 ) K

0.993 hrs ( 3600 s1hr )

H=76.68 kW

Thus, the total heat required for drying the raw materials is:

HTotal=76.65 kW +76.65 kW +76.68 kW

HTotal=229.98 kW

C. Grinder (Size reduction)

Assumptions:

1. There is no mass loss in the milling process.

2. Grinding process produces uniform particle size.

3. The work index used is from the smallest possible value for work index since

there is no available literature relating the work index to the cassava peels,

pineapple pomace and ipil- ipil leaves and that is from the clay material, which

has a work index of 6.30 (McCabe, Smith, & Harriott, 2005).

4. The reduction ratio for cassava peels and pineapple pomace, DP/DF, is 40 (Kayode

Coke, 2007) since there is no known size for the two raw materials.

5. The mass flow rate for all the three is at the maximum capacity, 2 kg/s.

The raw materials will be reduced to particle size 1 mm which will be aided by

the hammer mill. These mills are applicable for materials with solid hardness, 1-3 moh.

Size reduction is basically employed to improve the mixing efficiency of ingredients in

the biomass. It would also allow an easier access of the nutrients for consumption of the

yeasts.

47

Figure 3.3.2.5: Energy balance around Grinder for CP and PP.

The equation applicable for this type of mill is shown in equation 1 (McCabe,

Smith, & Harriott, 2005).

Pm

=0.3162W i( 1

√DP

−1

√DF) Eq. 95

where P=power∈kW

msmax=maximumcapacity ,

tonshr

DP=product ¿mm

DF=feed ¿mm

W i=work index of the material

For cassava peels and pineapple pomace:

From the assumed reduction ratio of 40 and a product size of 1 mm, the feed size is

calculated.

DP=Reduction ratio ( DF )DP=40 (1 mm )

DP=40 mm

Substituting the values,

Pm

=0.3162W i( 1

√DP

−1

√DF)

P

2kgs ( 3600 s

hr )( 1ton1000 kg )

=0.3162(6.30)( 1

√1mm− 1

√40 mm )P=12.075 kW

P=12 kW

Thus, the power used for each of the grinding process of cassava peels and

pineapple pomace is 12 kW.

For the ipil – ipil leaves:

48

Figure 3.3.2.6: Energy balance around grinder for IL.

The size of ipil – ipil leaves ranges from 2 – 5 cm (Stuart, 2014). Taking the mean

of the highest and lowest value, the assumed size of ipil – ipil leaves is 3.5 cm or 35 mm.

The same work index was used. The power needed for the grinding process is then

calculated.

Pm

=0.3162W i( 1

√DP

−1

√DF)

P

2kgs ( 3600 s

hr )( 1ton1000 kg )

=0.3162(6.30)( 1

√1mm− 1

√35 mm )P=11.918 kW

P=12 kW

D. Mixer 1

The mixer 1 is used to mix the substrates and the mineral solution with water

before it goes to the fermenter. It has 40% solids, 55% water, and 5% mineral solution as

its feeds. The schematic presentation of the system is shown below.

Figure 3.3.2.7: Energy balance around Mixer 1.

49

From the composition of the feeds, which is 40% solids and 60% liquid, in the

mixer the mixture at hand can be considered to be a light paste. It can be noted then that

mixing is difficult to define and evaluate with solids and pastes than it is with liquids as

in the previous activities. (McCabe, Smith, & Harriott)

Based on literature and other references, the density of the solids, composed of

cassava peels, pineapple pomace, and ipil-ipil leaves is ρ solids=256.6991kg /m3. Given

this information, the density of the slurry can then be determined by adopting the

calculation from EngineeringToolbox.com,

ρm= 100

[ cw

ρ s

+[100 – cw ]

ρl]

Eq. 123

where

ρm=density of slurry ( lb / ft3 , kg/m3)

cw=concentration of solidsby weight∈the slurry (%)

ρ s=density of the solids (lb / ft3 , kg /m3) and is equal to 256.6991 kg /m3

ρl=density of liquid without solids (lb / ft3 , kg /m3) (Slurry - Density)

Assumption:

1. The density of the liquid to be used is that of water since the mineral solution has

a negligible amount.

ρm= 100

[ 40

(256.6991kgm3 )

+[100 – 40 ]

999.98kgm3 ]

ρm=463.34kg

m3

The density of the mixture is 463.34 kg/m3. Thus, volume of slurry in the mixture

can be determined since the mass is given to be 23,606.09 kg,

50

ρ=mV

Eq. 124

V slurry=23,606.09 kg

463.34kg

m3

V slurry=50.95 m3

Typically, the working volume will be 70-80% of the total mixer volume. For the

study, a working volume of 70% is assumed.

V tank=50.95 m3

0.70=72.78 m3

It is predicted that the design of the mixer will be similar to that of a gate or

anchor paddle shown in Figure 2 below since it is applicable also for mixing of solids and

liquids. The paddle fills the tank completely and thus can scrape the solids stuck in the

walls of the tank.

Figure 3.3.2.8: Gate or anchor paddle(Geankoplis, 1993).

Since viscosity is needed to determine the Reynolds Number, Re, of the slurry,

an assumption or estimation is made. It is assumed that the slurry produced will have a

viscosity similar to that of ketchup or mustard. According to Viscosity Scales (Viscosity

Sales), the viscosity of peanut butter ranges from 150,000-150,000 cP as shown by the

Figure below.

51

Figure 3.3.2.9: Viscosity range of different products (Viscosity Sales).

The average of the range was taken to be used for the calculations.

μ=150,000 cP+250,000 cP2

μ=200,000 cP

Thus, the mixture has an estimated viscosity of 200,000 cP or 200 Pa∙s.

In order to calculate for the tank height and diameter, the following assumptions

are made.

Assumption:

1. The height and diameter of the tank is similar, H t=Dt .

52

V=π Dt

2 H4

=π Dt

2 ( Dt )4

V=π Dt

3

4

Eq. 125

Dt=3√ 4 V

π=

3√ 4(72.78 m3)π

Dt=4.52 m=H

For the agitator impeller diameter, it is said that Da ¿2/3 of tank diameter. Thus,

Da=23

Dt

Eq.126

Da=23(4.52m)

Da=3.01m

Since no reference can be used for this type of system, it is assumed that the

agitator rpm is 30 rev/min.

N=30revmin

=0.5revsec

Calculating for Reynolds Number, NRe, in order to determine the power number

NP using Figure 4 (Geankoplis, 1993).

N ℜ' =

Da2 Nρμ

Eq. 127

N ℜ' =

(3.01 m )2(0.5revsec )(463.34

kg

m3 )200 Pa ∙ s

N ℜ' =10.49≈ 10

A red line is made to locate the curve and power number as shown in Figure 4

below.

53

Figure 3.3.2.10: Power correlations of various impellers and baffles (Geankoplis, 1993).

Since the impeller used is that of the gate paddle, the curve to be used here is only

assumed. It is assumed that Curve 4 is similar to the one used.

Thus, from the figure, the system has a power number of approximately 4.

Substituting to the equation to solve power requirement,

N P=P

ρ N 3 Da5∨P=NP ρ N3 Da

5 Eq. 128

P= (5 )(463.34kgm3 )(0.5

revs )

3

(3.01m )5

P=76,432.67Js∨W =76.43 kW

P=76.43 kW ( 1 hp0.7475 kW )=102.25 hp

∴The power requirement needed for operating this mixer is 102.25 hp .

E. Mixer 2

54

The mixer 2 is used to mix water with the minerals in order to produce mineral

solution that is needed for the microorganisms to grow. From the material balance, the

solid component is just 5.6% of the total mixture while the liquid component, or water,

constitute to the 94.4%.

Since, the mixture in mixer 2 is mainly composed of water, the following

assumptions are made in order to compute for the power.

Assumption:

1. The density and viscosity of the mixture is that of water since the mineral solution

has a negligible amount. Thus, ρmixture = 999.98 kg/m3 and µ = 0.001 Pa·s

From the material balance, the total amount of mixture in the mixer is 1168.50 kg,

hence, the volume of the mixture can be determined.

ρ=mV

Eq. 124

V mineral soln=1168.50kg

999.98kg

m3

V mineral soln=1.17 m3

Typically, the working volume will be 70-80% of the total mixer volume. For the

study, a working volume of 70% is assumed.

V tank=1.17 m3

0.70=1.67m3

It is predicted that the design of the mixer will be similar to that of a flat six –

blade turbine with disk having four baffles each.

In order to calculate for the tank height and diameter, the following assumptions

are made.

Assumption:

1. The height and diameter of the tank is similar, H t=Dt .

55

2. The ratio of the agitator diameter to the tank diameter is 0.4. This is the average

geometric proportion from the range given in Table 3.4-1 in Geankoplis, 1993.

V=π Dt

2 H4

=π Dt

2 ( Dt )4

V=π Dt

3

4

Eq. 125

Dt=3√ 4 V

π=

3√ 4(1.67 m3)π

Dt=1.29 m=H

For the agitator impeller diameter, it is said that Da ¿0.4 of tank diameter. Thus,

Da=0.4 Dt Eq.126

Da=0.4(1.29 m)

Da=0.516m

Since no reference can be used for this type of system, it is assumed that the

agitator rpm is 30 rev/min.

N=30revmin

=0.5revsec

Calculating for Reynolds Number, NRe, in order to determine the power number

NP using Figure 4 (Geankoplis, 1993).

N ℜ' =

Da2 Nρμ

Eq. 127

N ℜ' =

(0.516 m)2(0.5revsec )(999.98

kg

m3 )0.001Pa ∙ s

N ℜ' =133,125.34

A red line is made to determine the curve and power number in Figure 3.3.2.10.

Since the impeller used is that of flat six – blade turbine with disk having four baffles

each. It is said to be that Curve 1 is the one to be used. From the figure, the system has a

56

power number of approximately 4. Substituting to the equation to solve power

requirement,

N P=P

ρ N 3 Da5∨P=NP ρ N3 Da

5 Eq. 128

P= (5 )(999.98kgm3 )(0.5

revs )

3

(0.516 m )5

P=22.86Js∨W =0.023 kW

P=0.023 kW ( 1 hp0.7475 kW )=0.03 hp

∴The power requirement needed for operating this mixer is 0.03 hp .

F. Fermentor

The fermentation process is shown below:

Figure 3.3.2.11: Energy balance around Fermentor.

Assumptions:

1. Heat for combustion for yeast is -21.2 kJ/kg (Doran, 1995). This is because at the

present time, the thermodynamic properties of cells cannot be found in literature

because cells are not pure substances. Cells also don’t have finite enthalpy,

entropy and free energy of formation. In order to obtain the cellular enthalpy of

formation, it is necessary to measure the heat of combustion and to construct an

equation representing the combustion of a unit mass of cells (Kemp, 1999).

57

2. Agitation is minimal. Thus Ws = 0.

3. Evaporation is negligible. Thus, Mv = 0.

4. All the N-content of ipil – ipil leaves and in the mineral solution are consumed in

the process as a source of NH3.

5. There is glycerol and water formed during the reaction, however, for glycerol,

ther was no literature available relating the production of biomass from the

substrates to the glycerol formed. Thus, it is assumed that the amount of glycerol

formed is minimal and negligible. For the water formed, it is assumed that the

change in the water content in the fermentation tank is negligible since during

fermentation, a large amount of water is added.

6. In the energy balance, only the glucose in the substrate is reacted and is accounted

and the biomass stated is the accumulated protein.

The chemical reaction for the fermentation process is given below:

Substrate+N H 3 → Biomass+Glycerol+Ethanol+C O2+H 2O

Since negligible glycerol and water produced, the chemical reaction becomes:

GlucoseSubstrate+ N H 3 → Biomass+C O2+Ethanol Eqn. *

In order to determine the amount of glucose present in the substrate, the glucose

in the cassava peels and pineapple pomace are calculated. From the material balance, it is

known that the feed that is introduced in each fermenter is 394 kg and contains 137.55 kg

cassava peels and 137.55 kg pineapple pomace. Hence, the amount of glucose in each

fermenter is calculated.

For the glucose in the cassava peels (CP), according to literature, cassava peels

contain 51.93% starch. Thus, the amount of starch is given by:

mstarch=mCP∈ substrate (% starch)

mstarch=4,089.76 kgCP( 0.5193 kg starchkgCP )

mstarch=2123.81 kg starch

In order to calculate the amount of glucose from starch, the equation is given by:

58

mglucose=mstarch

0.9

Thus, the amount of glucose can be calculated below:

mglucose=2123.81 kg starch

0.9

mglucose=2359.79 kg glucose

For the glucose in the pineapple pomace (PP), according to Correia, Magalhães,

& Macêdo, 2007, pineapple pomace contain approximately 27.2% glucose. Thus, the

amount of glucose from the pineapple pomace can be calculated and is given by the

equation below:

mglucose=mPP∈ substrate (% glucose )

mglucose=4,089.76 kg PP( 0.272 kg glucosekg PP )


Adding the two values to calculate the total amount of glucose present in the

substrate:

mtotal glucose=mglucose ¿CP¿+mglucose ¿PP ¿mtotal glucose=2359.79 kg glucose+1112.41 kg glucose

mtotal glucose=3472.20 kg glucose

For the amount of NH3 in the fermentation process, it is assumed that the NH3 is

obtained from three different sources: ipil – ipil leaves, cassava peels and mineral

solution. According to Abdullah & Hossain, 2006, ipil – ipil leaves contain 4.3% nitrogen

(N) and from Ismadji, Ju, Chun, Kurniawan, & Ong, 2012, cassava peels contain 1.2% N.

Also, in the mixer 1, a mass of 1.179 kg of (NH4)2SO4 is added.

Hence, in order to calculate the number of moles of NH3, it is given by:

For the NH3 in CP:

nN H 3=mCP (% N )( 1

MW of N )( 1mol N H 3

1mol N )nN H 3

=4,089.76 kgCP( 0.012kg NkgCP )( 1000 g

1 kg )( 1 mol N14.01 gN )(1 mol N H 3

1 mol N )nN H 3

=3503.01 mol N H 3

For the NH3 in IL:

59

nN H 3=mIL (% N )( 1

MW of N )(1mol N H 3

1mol N )nN H 3

=1,168.50kg IL( 0.043kg Nkg IL )( 1000g

1kg )( 1mol N14.01gN )( 1mol N H3

1mol N )nN H 3

=3586.40 mol N H 3

For the NH3 in the mineral solution:

nN H 3=m( NH 4 ) 2SO 4( 1000g

1kg )( 1MW of ( NH 4 )2SO 4 )( 2mol N

1mol ( NH 4 ) 2SO 4 )( 1mol N H 3

1mol N )nN H 3

=35.06 kg (1000 g1kg )( 1 mol ( NH 4 ) 2 SO 4

32.065 g ( NH 4 ) 2 SO 4 )( 2 mol N1mol ( NH 4 ) 2SO 4 )( 1mol N H 3

1 mol N )nN H 3

=2186.81 mol N H 3

Adding the three values:

nN H 3=3503.01 mol+3586.40 mol+2186.81 mol

nN H 3=9276.22 mol N H 3

The mass of the NH3 produced is equal to:

mN H 3=9276.22 mol N H 3( 17.034 g

1mol )( 1kg1000 g )

mN H 3=158.01 kg N H 3

In the fermentation process, it is assumed that not all glucose in the substrate is

consumed. From the material balance, it is determined that the amount of accumulated

biomass, ethanol and carbon dioxide are 118.03 kg, 542.94 kg and 542.94 kg

respectively. Hence, the amount of glucose consumed is calculated using the material

balance around the fermenter and is given by:

m¿=mout

mglucose+mN H 3=mbiomass+methanol+mC O2

mglucose+158.01 kg=118.03kg+542.94 kg+542.94 kg


In order to calculate the percent of glucose reacted or consumed, it is given by:

% glucose reacted=mglucose consumed

mtotal glucose

×100%

60

% glucose reacted=1045.90 kg glucose3472.20 kg glucose

×100 %

% glucose reacted=30.12 %

For the energy balance for the fermentation process, it is adapted in the book of Doran

and is given below:

−∆ H rxn−M v hv−Q+W s=0 Eq. 98

Since Mv = 0 and Ws = 0, the equation becomes:

−∆ H rxn−Q=0

−∆ H rxn=Q

The ΔHrxnis the heat of combustion reaction for the chemical reaction below:

Glucose+N H 3→ Biomass+Ethanol+C O2

Thus, the heat of combustion is given below:

∆ H rxn=m(∆ hc)products−m(∆ hc )reactants Eq. 99

The heats of combustion in the fermentation process are given below:

( ∆ hc )glucose=(−2,805.0kJ

gmol )( 1 gmol180 g )( 1,000 g

kg )=−15,583.33kJkg

( ∆ hc )N H 3

=(−382.6kJ

gmol )( 1gmol17 g )( 1,000 g

kg )=−22,505.88kJkg

( ∆ hc )ethanol=(−1,655.4kJ

gmol )( 1 gmol92 g )( 1,000 g

kg )=−17,993.48kJkg

( ∆ hc )protein/ yeast=21.2kJkg

( ∆ hc )carbondioxide=0

Thus, the heat of reaction can be calculated below:

∆ H rxn=m(∆ hc)products−m(∆ hc )reactants

mglucose+158.01 kg=118.03kg+542.94 kg+542.94 kg

∆ H rxn=[118.03kg (−21.2kJkg )+542.94kg ethanol(−17,993.48

kJkg )+542.94kgC O2 (0 )]−[1045.90 kgglucose (−15,583.33

kJkg )+158.01kgN H3(−22,505.88

kJkg )]

61

∆ H rxn=10,082,876.68 kJ

Thus, Q = −10,082,876.68kJ for each fermenter. This means that the heat should

be removed in the fermentation. Thus, cooling water must be supplied during

fermentation to maintain the condition at 30°C. The equation is given by:

Q=mcooling water c p ∆T Eq. 100

From heuristics, it is found out that the maximum ΔT for cooling water is 30°F. It

is assumed that the temperature of the entering cooling water is 20°C ≈ 68°F and the

temperature of the exiting cooling water is the same as the fermenter’s temperature,

which is 30ºC ≈ 86°F. Hence, it gives a ΔT of 10°C / K or 18°F. Also, from the table of

the properties of water, it is obtained that the cp,water = 4.187 kJ/kg·K using interpolation

at the temperature of 20ºC. Thus, the mass of the cooling water can now be computed.

Q=mcooling water c p ∆T

10,082,876.68 kJ=mcooling water (4.187kJ

kg·K )(10 K )

mcooling water=240,813.87 kg cooling water

Therefore, for each fermenter in one batch, the amount of cooling water needed is

240,813.87 kg. On an hourly basis, it can be computed since one batch of fermentation

would consume approximately 3 days.

mcooling water=240,813.87 kgcooling water

batch ( 1 batch3 days )( 1 day

24 hours )mcooling water=3344.64

kghr

cooling water

mcooling water=3344.64kghr

cooling water ( 1 hr60 min )( 1 m3

999.97 kg )( 1000 L1m3 )( 1gal

3.785412 L )mcooling water=14.73 gpm cooling wa ter

K. Filter Press

62

For the solid – liquid separation of the product produced after fermentation, a filter

press is used. There is no definite equation to solve the power requirement for the filter

press. Thus, the power requirement is achieved through a design specification. The filter

press to be used is Outotec Larox FFP 2 512. From the design specification, it is desired

to separate a filter volume of 14.29 m3 and the equipment has a filter volume range of 5.4

– 15.4 m3, that is why, the equipment is appropriate for the desired solid – liquid

separation. Hence, for this type of equipment, the installed power (hydraulics) is 90 kW,

which is given from the design specification (Outotec Larox, 2013).

L. Pelletizer

In pelletization, biomass is compressed against a heated metal plate (known as die)

using a roller. The die consists of holes of fixed diameter through which the biomass

passes under high pressure. Due to the high pressure, frictional forces increase, leading to

a considerable rise in temperature. High temperature causes the lignin and resins present

in biomass to soften which acts as a binding agent between the biomass fibers. This way

the biomass particles fuse to form pellets.

The rate of production and electrical energy used in the pelletization of biomass are

strongly correlated to the raw material type and processing conditions such as moisture

content and feed size. The average energy required to pelletize biomass is roughly

between 16 kWh/t and 49kWh/t. During pelletization, a large fraction of the process

energy is used to make the biomass flow into the inlets of the press channels. (Zafar,

2014)

Assumptions:

1. The pelletizer applicable to this process operates between 16 kWh/t to 49kWh/t.

Solution:

Take average energy of pelletizer,

E=(16+49 ) kWh / t

2

E=32.5 kWh /t

63

Multiplying the amount of feed (from material balance) injected to the pelletizer,

Energy requirement=32.5kWh

t ( 1 ton1000 kg ) (10,000 kg )

Energy requirement=325 kWh

In the pelletizer, it is specified to have a total time of 1 hour in its operation. Thus,

calculating the energy requirement in kW:

∴Energy requirement=325 kWh (1 hr)

∴Energy requirement=325 kW

M. Tunnel Dryer

In drying the final output after filter press, the pellets are discharged into a screen

belt of a horizontal tunnel drier. In this dryer, the materials to be dried are sent to the air

heated tunnel for drying purpose. The material will enter at one end and the dried

material is collected at the other end of the tunnel. The outgoing material meets the

incoming air to ensure maximum drying and the outgoing air contacts the wettest

material so that the air was as nearly saturated as possible (Islam, 2012).

The tunnel dryer used has a 1.2 m width, 8 m drying section length and a 10 – 80

mm thickness of material covered. It operates at a temperature of 60 – 100°C having a

steam pressure of 0.2 – 0.8 MPa. The steam consumption is 120 – 300 kg/h. The power

required for the blower is 9.9 kW and the power required for the equipment is 11.4 kW

(Zhengzhou Bangke Machinery Manufacturing Co., Ltd., 2014).

ii. Utilities

a. Pumps

Cooling Water Pump

Assumptions:

1. A 2 – in. Schedule 40 nominal size for the pipe was used.

2. An elevation of 50 ft was used.

3. The efficiency of the pump is 40%.

64

From the energy balance around the fermenter, the volume flow rate of the

cooling water is obtained. It has a value equal to 31.52 gpm. Converting the value into a

unit of ft3/s:

VFR=14.73galmin

×1 min60 sec

×1 ft3

7.481 gal

VFR=0.0328ft3

s

From the assumed size of the pipe, the cross – sectional area of the pipe is

0.02330 ft2. Having the cross – sectional area, the velocity of the cooling water is given

by:

v=VFRA

v=0.0328

ft3

s0.02330 ft2

v=1.41fts

Using the overall mechanical energy balance:

v2

2 gc

+∆ zggc

+ ∆ pρ

+∑ F=−W s

Eq. 101

For the kinetic energy:

v2

2 gc

=(1.41

fts )

2

2(32.174lbm ∙ ftlbf ∙ s2 )

v2

2 gc

=0.031lbf ∙ ftlbm

For the potential energy:

∆ zggc

=50 ft ( 32.174ft

s2

32.174lbm ∙ ftlbf ∙ s2 )

65

∆ zggc

=50lbf ∙ ftlbm

For the friction losses inside the pipe:

For the friction loss in pipe fittings:

Assumptions:

1. Use 1 – 90° elbow

2. Use 1 gate valve

The Kf for a 90° elbow is equal to 0.75. Hence, the friction loss for the 90° elbow

is calculated.

h f=K f ( v2

2 gc ) Eq. 102

h f=0.75[ (1.41fts )

2

2(32.174lbm ∙ ftlbf ∙ s2 ) ]

h f=0.023lbf ∙ ftlbm

The Kf for the gate valve is 0.17. Thus, the friction loss due to the use of gate

valve is obtained.

h f=K f ( v2

2 gc )

h f=0.17[ (3.014fts )

2

2(32.174lbm ∙ ftlbf ∙ s2 ) ]

h f=0.005lbf ∙ ftlbm

For the friction loss in the 2 in. pipe:

Assumption:

1. Commercial steel pipe is used.

66

2. The length of pipe is 150 ft.

Calculating the Reynolds number for the cooling water:

N ℜ=Dvρ

μEq.103

At a temperature of 25°C, the density of water is 62.2477 lbm /ft3, its viscosity is

6.00539589x10-4lbm / ft·s and the inner diameter of the pipe is 2.067 in

(Geankoplis, 1993). Thus, the Reynolds number is obtained.

N ℜ=Dvρ

μ

N ℜ=(2.067

12ft)(1.41

fts )(62.2477

lbm

ft3 )6.00539589 ×10−4 lb m

ft·s

N ℜ=25,174.45

For a commercial steel pipe, the equivalent roughness is equal to 4.6x10-5m or

1.509186352x10-4-ft (Geankoplis, 1993). Computing the relative roughness:

εD

=1.509186352 ×10−4 ft2.067

12ft

εD

=8.7616 ×10−4=0.00087616

For an NRe of 25,174.45 and a relative roughness of 0.00087616, the friction

factor from Fig. 2.10-3 is f = 0.006. The friction loss is calculated and is given by:

F f =4 f∆ LD

v2

2 gc

Eq. 104

F f =4 (0.006) 150 ft

( 2.06712

ft )(1.41

fts )

2

2(32.174lbm ∙ ftlbf ∙ s2 )

67

F f =0.646lbf ∙ ftlbm

Thus, the total friction loss can be calculated.

∑ F=0.023lbf ∙ ftlbm

+0.005lbf ∙ ftlbm

+0.646lbf ∙ ftlbm

∑ F=0.674lb f ∙ ft

lbm

For the pressure difference:

In order to calculate for the pressure difference, the pressure head was used. A

table is found in Static Pressure and Pressure Head in Fluids, 2014 where velocity, in ft/s,

corresponds to the certain pressure head of water, in ft. The table is shown below:

Since the calculated velocity is 1.41 ft/s, the value of the pressure head is

interpolated from the table given. The value for the pressure head of water is 0.03158 ft.

The pressure difference is calculated using the equation below:

p2−p1=ρghgc

Eq. 105

∆ p=(62.2477

lbm

ft3 )(32.174ft

s2 )(0.03158 ft )

32.174lbm∙ ftlbf ∙ s2

∆ p=1.966lb f

f t2

The shaft work for the pump can be calculated.

v2

2 gc

+∆ zggc

+ ∆ pρ

+∑ F=−W s

68

0.031lbf ∙ ftlbm

+50lbf ∙ ftlbm

+1.966

lbf

f t2

62.2477lb mft3

+0.674lbf ∙ ftlbm

=−W s

−W s=50.74lbf ∙ ftlbm

Calculating the mass flow rate of the cooling water:

m=VFR × ρ

m=0.0328ft 3

s×62.2477

lbm

ft3

m=2.04lbm

s

Calculating the shaft work in lbf /ft·s:

−W s=50.74lbf ∙ ftlbm

× 2.04lbm

s

−W s=103.51lbf ∙ ft

s

Calculating the work needed for the pump:

W p=−W s

η

W p=103.51

lbf ∙ fts

0.40

W p=258.77lbf ∙ ft

s

Thus, the work can be calculated in hp.

W p=258.77lbf ∙ ft

s×

1 hp

550lbf ∙ ft

s

W p=0.47 hp=0.35kW

Slurry Pumps (for slurry materials)

Pumping slurry materials is more difficult than pumping liquid materials. One

type of pump that aids the pumping of slurry is slurry pump. These are which are capable

69

of handling tough and abrasive materials. The determination of the power requirement of

the slurry pump is constrained by the limited data available for the pump. However, it

was determined that the standard power requirement for slurry pumps having a RPM of

1455 is 16 kW. A total of eight (8) slurry pumps will be used in the system. (Filter

Sterilization Guide: Steam Sterilization & Alternative Methods)

b. Screw Conveyor

hp=10−6( ALN +CWLF) Eq.

where A=sizefactor=54

L=conveyorlength

N=conveyor(rev /min)= (154 mm ) 100 rev /min

C=quantityofmaterialhandled=10,000 kg∨410.056 f t3

W =densityofmaterial=53.764 lb / f t 3

F=materialfactor=0.5

ɳ=90 %efficiency

Assumption: L=50 m∨62500 /381 f t

hp=10−6 [(54 )( 62500381

ft)(100revmin )+ (410.056 f t3 )(53.764

lbf t3 )( 62500

381ft ) (0.5 )]

hp=2.694 hp

hp ≈ 2.694 hpor2.01 kW

With respect to efficiency,

W required=Wɳ

W required=2.694 hp

0.9

W required=2.99 hp∨2.23 kW

c. Belt Conveyor

Discharge Belt Conveyor in the Washer

70

Assumption:

Length: 5-20 feet, assumed to be 15 ft

hp=(C /100)(0.4+0.00345 L)

where L=distanceconveyor centers (ft )=15 ft

C=quantity of material=6,252.78 kg∨6.253ton

W =width of belt conveyor=30∈.

Empty: 1.2 hp

To move the load: 0.10 hp

Thus,

Total hp=1.2 hp+0.028 hp

Total hp=1.228 hp=0.916 kW

d. Air Heater

For the air heater, the following equation is used to calculate for the required

heating capacity (Chromalox Precision Heat and Control).

H=H Total × SF Eq. 106

where: HTotal = total heat required for drying

SF = safety factor

For this, it is assumed that the heater has a safety factor of 20%. The total heat required

for drying is 412.61 kW. Thus, the heating capacity can be obtained.

H=H Total × SF

H=229.98 kW ×1.2

H=275.98 kW

e. Compressor

For the Air for Drying:

Assumption:

1. p2 = 151.99 kPa and p1 = 95.36 kPa

2. An adiabatic compression is assumed.

71

3. Efficiency, η = 0.70

The mass of total air is expressed in kg/s using a drying time of 1 day or 24 hrs

and is shown below:

m=210,028.54 kg48 hrs

×1 hr

3600 s

m=1.215kgs

Calculating for the heat capacity of air having a humidity of 0.05 kg water

vapor/kg dry air:

c p=1.005+1.88 (H )

c p=1.005+1.88 (0.05 )

c p=1.099kJ

kgdry air ∙K

The heat capacity is expressed in J/mol·K using a molecular weight of air to be 29

kg/kgmol.

c p=1.099kJ

kgdry air ∙K×29

kgkgmol

c p=31.871kJ

kgmol ∙ K

The ratio of heat capacities, γ, is then calculated.

γ=c p

cv

=c p

c p−R

Eq. 107

γ=31.871

kJkgmol ∙ K

31.871kJ

kgmol ∙ K−8314.34

Jkgmol ∙ K ( 1kJ

1000 J )γ=1.353

Thus, the shaft work can be calculated.

−W s=γ

γ−1R T1

M [( p2

p1)

γ−1γ −1] Eq. 108

72

−W s=1.353

1.353−1

(8314.34J

kgmol ∙ K ) (120+273.15 ) K

29kg

kgmol[(151.99 kPa

95.36 kPa )1.353−1

1.353 −1]−W s=55,872.25

Jkg

Calculating the brake power:

brake kW =−W s m

η× 1000

brake kW =55,872.25

Jkg (1.215

kgs )

0.70×1000J

kJ

brake kW =96.98 kW

It can also be expressed as brake hp and is given by:

brake hp=96.98 kW ×1 hp

0.74570 kW

brake hp=130.05 hp

f. Blower (for Washer)

Assumptions:

1. Air flow rate is 1/2 of mass flowrate of water (10 m3/hr).

2. Standard air density: 1.2007 kg/m3

3. Adiabatic compression occurs.

4. Mechanical efficiency of blower is 70%.

5. γ=1.40 for air (Geankoplis, 1993).

6. M=28.97kg

kmol for air.

Calculation:

mair=12

mH2 O

=12 (10,000

kghr )=5,000

kghr

=1.39kgs

Equation to use,

73

−W s=γ

γ−1RT1

M [( p2

p1)

γ−1γ −1] (Eqn. *)

Determine p1 (absolute initial pressure).

Manolo Fortich is located 509 m above sea level (www.maps-streetview.com,

2011), calculating for initial pressure (atmospheric),

p1=101,325 (1−2.25577 x10−5 h )5.25588

p1=101,325¿

p1=95,357.86 Pa

Determine p2 (absolute final pressure).

Assume p2 to be 101.325 kPa.

Substitute known values to equation above,

−W s=1.40

1.40−1 ((8,314.3J

kmol ∙ K )(298.15 K )

28.97kg

kmol) [(101.325 kPa

95.36 kPa )1.40−1

1.40 −1]−W s=5,237.02

Jkg

Calculate power requirement (brake kW).

brake kW =−W s m

ɳ ∙ 1000

(Eqn. *)

brake kW =(5,237.02

Jkg )(1.39

kgs )

0.70(1000)

brake kW =10.39 kW (13.93 hp)

3.4. Equipment Sizing and Specification

The section aims to present the equipment design and specification for the various

equipments involved in the production of high-value animal feeds from agro-industrial

wastes.

74

a. Dryer

For the study, two sets of drying procedure are to be conducted. The first drying is

done prior to fermentation. The second drying is then done in order to reduce the

moisture content of the final product to the desired value. The equipment specification of

the report will focus on the drying procedure of the raw materials prior to processing.

Moreover, the drying process is employed in batch condition. The figures below illustrate

the schematic diagrams of the drying procedure for each raw material used.

Figure 3.4.1: Schematic Diagram of the drying procedure for cassava peels.

Figure 3.4.2: Schematic Diagram of the drying procedure for pineapple pomace

75

Figure 3.4.3: Schematic Diagram of the drying procedure for ipil-ipil leaves

The drying procedure employed in the production of animal feeds would involve

the use of tray dryer as the drying equipment. In a tray dryer, the material is uniformly

spread on a metal tray. Heated air is then used as the heat-source for the system.

The material and energy balance are considered in the calculations of the design

parameters of the dryer. The data for this is given in Section 2 and 3 of this paper.

Moreover, for the equipment specification, the drying of the pineapple pomace is system

of interest. For the calculations, the following conditions are adopted:

Table 3.4.1: Drying Parameters

Drying Parameters Values

Inlet Temperature of air (°C) 120

Outlet Temperature of air (°C) 57.7441

Inlet Temperature of solid (°C) 25

Outlet Temperature of solid (°C) 60

Inlet Humidity of air (kg moisture/kg DA) 0.05

Outlet Humidity of air (kg moisture/kg DA) 0.075

i. Calculating for humid volume, vH:

76

Using equation (Geankoplis 2003) given below, the humid volume can be

calculated.

vH=(2.83×10−3+4.56 ×10−3 H )T Eq. 109


vH=(2.83× 10−3+4.56× 10−3 (0.05 ) ) (120+273 )

vH=1.2018m3

kg DA

ii. Calculating for the density of the air entering the system:

ρ=(1+0.05 )1.2018

ρ=0.8737kg

m3

iii. Calculating for the mass velocity of air:

Assuming that the velocity of the air entering between trays is 2 m/s. the mass

velocity can then be calculated.

G=V i ρ Eq. 110

G=(2 ms )(0.8737

kg

m3 )G=1.7474

kg

m2 ∙ s

iv. Calculating for the heat transfer coefficient:

The heat transfer coefficient can be calculated using the equation (Geankoplis

2003) given below:

h=0.0204 G0.8 Eq. 111


77

h=0.0204 (1.7474kg

m2 ∙ s×

3600 s1hr )

0.8

h=22.31W

K ∙ m2

v. Calculating for the area between trays:

For the calculations, it is assumed that there are 24 trays available. Based on

energy balance, the amount of dry air is 210,028.54 kg. The area between the trays can

then be calculated using the equation (Geankoplis 2003):

Abetween trays=amount of air enterin g per second per tray

air density × velocity of air enteringbetweentraysEq. 112


Abetween trays=

210,028.54 kgdry air4824

kgh ∙ tray

×1 h

3600 s

0.8737kg

m3 ×2ms

Abetween trays=0.02898m2

tray≈ 0.03

m2

tray

vi. Calculating for the width of the trays:

It is assumed that the distance between the trays is 3 cm. The width of the trays

can then be calculated.

W trays=Abetween trays

distance betweentrays

Eq. 113

W trays=0.03

m2

tray0.03 m

78

W trays=1.00 m

vii. Calculating for the length of the trays

The calculation for the length of the trays was employed using the equation

(Geankoplis 2003):

H Lt

GC s (b W trays )=ln ¿¿

Eq. 114

It is known that the wet bulb temperature of the system is 48°C. Substituting known

values:

Lt=G C s ( b W trays )

hln ¿¿

Eq. 114

where C s=( (1.005+1.88 (0.05 ) ) ×1000 )=1,099

Substitute to Eq. *,

Lt=(1.7474kg

m2 ∙ s )(1,099)¿¿

Lt=5.1647 m≈ 5.16 m

ix. Calculating for the depth of the tray:

In the calculation for the depth of the tray, the density of the material to be dried

is to be considered. For the system, the density of the pineapple pomace, cassava peels

and ipil-ipil leaves are 150, 862.5 and 265 kg/m3, respectively. 7,431.21 kg of pineapple

pomace, 5,684.35 kg cassava peels and 1,240.63 kg ipil-ipil leaves are introduced to the

drier. Moreover, it is assumed that two batches for drying will be employed. The volume

of the pineapple pomace per tray can be calculated.

Pineapple pomace:

79

V pineapple pomace per tray=

7,431.21 kg2

24 trays per batch×

1

150kg

m3

V pineapple pomace per tray=1.0321m3

Cassava peels:

V cassava peels per tray=

5,684.35kg2


1

862.5kg

m3

V cassava peels per tray=0.1373 m3

Ipil-ipil leaves:

V ipil−ipil leaves per tray=

1,240.63 kg2


1

265kg

m3

V ipil−ipil leaves per tray=0.0975 m3

The volume of the tray can then be calculated (adopting the conditions for pineapple

pomace). The tray is assumed to be 80% full, then, the tray volume is:

V tray=1.0321 m3

0.80

V tray=1.2901 m3

Therefore, the depth of the tray is:

Dtray=V tray

Ltray× W tray

Eq. 115

80

Dtray=1.2901 m3

5.16 m× 1 m

Dtray=0.2500 m≈ 0.25 m=25 cm

Then, the depth of each material:

Dmaterial=V material

Ltray× W tray

Eq. 116

Pineapple Pomace:

D pineapple pomac e=1.0321 m3

5.16 m×1m

D pineapple pomace=0.2000 m≈ 0.20 m

Cassava peels:

Dcassava peels=0.1373 m3

5.16 m×1 m

Dcassava peels=0.0266 m ≈ 0.03 m

Ipil-ipil leaves:

Dipil−ipil leaves=0.0975 m3

5.16 m ×1m

Dipil−ipil leaves=0.0190 m≈ 0.02m

x. Calculating for the height of the dryer:

It is assumed that 0.15 m is the distance of the bottom of the drier to the nearest

tray and 0.15 is also the distance of the top of the drier to the nearest tray. It is assumed

that the distance between the trays is 0.03 m. The height of the drier can be calculated

using the equation:

81

H dryer=(no .of trays ×depth of the trays )+(11× distancebetweenthe trays )+0.15(2)Eq. 117

H dryer=(24 × 0.25m )+(11× 0.03 m )+0.30 m

H dryer=6.63 m

xi. Calculating for the length of the drier:

It is assumed that 0.15 m is the distance of the left side of the drier to the tray and 0.15

is also the distance of the right side of the drier to the tray. There is also an existing

distance between the two columns equal to 0.15. The length of the drier can be calculated

using the equation:

Ldryer=( Number of columns× width of trays )+0.15+0.15+0.15 Eq. 118

Ldryer=(2× 1m)+0.15+0.15+0.15

Ldryer=2.45m

xii. Calculating for the width of the drier:

It is assumed that there are existing spaces around the dryer equal to 0.05 m. The

width can be calculated:

W dryer=length of trays+0.05+0.05 Eq. 119

W dryer=2.45 m+0.05+0.05

W dryer=2.55 m

xiii. Calculating for the area and diameter of the air inlet pipe:

The area of the bulk air in the inlet considering a bulk velocity of inlet air equal to 5

m/s:

Abulk airinlet=210,028.54 kgair

48 hr×

1

5ms

×1

1.7474kg

m2 ∙ s

×1 hr

3600 s

82

Abulk airinlet=0.1391m2 ≈ 0.14 m2

Then, the diameter of the pipe can be calculated:

Dbulk air inlet=( 4π

× A )12 Eq. 120

Dbulk air inlet=( 4π

× 0.14m2)12

Dbulk air inlet=0.4222 m ≈ 0.42m

Table 3.4.2: Equipment Specification Sheet for Tray Dryer

EQUIPMENT SPECIFICATION SHEETItem Name Tray DrierQuantity 1Project Description (i.e. pilot scale) Industrial ScaleFunction (i.e. reaction vessel) DrierMode of Operation (i.e. batch) Batch

TANK PARAMETERSOrientation (i.e. horizontal) Vertical Dimensions: height, length, tangent-to-

tangent, mLength = 2.45 m, Height = 2.45 m, Width

= 2.55 mDesign Temperature, ⁰C >300°CDesign Pressure, atm >10 atm

MATERIALS OF CONSTRUCTIONTank (i.e. Stainless Steel 316) Stainless steelTrays (i.e. Stainless Steel 316) Stainless steel

Jacket Type (i.e. simple, no baffles)FEED CONDITIONS

Inlet Temperature of air (°C) 120Inlet Temperature of solid (°C) 25Inlet Humidity of air (kg moisture/kg DA)

0.05

OUTPUT CONDITIONSOutlet Temperature of air (°C) 57.7441Outlet Temperature of solid (°C) 60Outlet Humidity of air (kg moisture/kg DA)

0.075

TANK CAPACITYVolume of drier, m3 41.42

83

b. Electric Heater

For the design specification of the electric heater, the following conditions are adopted:

Table 3.4.3: Electric Heater ConditionsInlet Outlet

Temperature (°C) 30 120

Gauge Pressure (atm) 5 -

Energy Required (kW) 18.4

Air Velocity (m/min) 90

Using the given conditions, the calculation was carried out.

i. Calculation of the air flow inside the duct.

Using the equation taken from Tempco Guidebook to Duct forced air heater

design, expressed as:

kW =CMM∗Density ( kg

m3 )∗∆T

57.5

Eq. 121

Rearranging,

CMM= 57.5∗kWDensity∗∆ T

= 57.5∗18.40.8737∗(120−30)

=13.45m3

min

ii. Calculation of Standard cubic meter per minute

SCMM=

13.45∗P1atm

∗274.15+30

T+274.15

where P = operating pressure, T = operating temperature (set at 121°C)

84

SCMM=

13.45∗5+11

∗30+273.15

121+273.15=62.09

m3

min

iii. Calculation of duct area.

The air velocity was calculated using the equation given below:

Air velocity= SCMMDuct crosssectional area

Eq. 122

Values for the air velocity inside the duct air heater system are obtained from Tempco’s

guidebook to designing air heaters.

Rearranging this equation will give the following result:

Are aduct=62.09

m3

min

90m

min

=0.69m2

Fixing one side of the duct to be equal to 0.5m, the width, the other side will simply be

the quotient of the area and the set dimension. Thus,

L=0.69 m2

0.5 m=1.38 m

iv. Estimating the pressure drop of the air inside the duct.

To estimate the pressure drop of the fluid traveling inside the duct the following

diagram, taken from the guidebook, is used below.

85

Figure 3.4.4: Air Velocity versus Approximate Pressure Drop Graph

The pressure drop is estimated by intersecting the air velocity value with that of

the energy requirement operating line of the given system. A line going down the X-axis,

the pressure drop axis, is then made starting from the intersection formed and the value of

the pressure drop is obtained. It is found that the pressure drop is at 0.00135 kPa.

Table 3.4.5: Equipment Specification Sheet for Air Duct Heaters

Operating Conditions 5. ELEMENT DIAMETER: 11 mm

1. APPLICATION (Describe in Detail): 6. TERMINAL SEALS:Silicone Fluid (260 C)

This air duct heater is use to heat air that will be use as the drying fluid for the drying operations of the cassava peel,

pineapple pomace, ipil-ipil leaves and the final animal feed product in the animal feed production.

7.TERMINAL BOX CONSTRUCTION:

General Purpose

Moisture Resistant

Explosion Resistant

2.AIR FLOW: 62.09 cu.m/min 8. TERMINAL BOX MATERIAL: 316 SS

3.INLET AIR TEMPERATURE: 30 °C 9.FLANGE MATERIAL: 304 SS

4.OUTLET AIR TEMPERATURE: 120 °C 10. INSULATION HOUSING (Below Flange):

5.OPERATING TEMPERATURE: 121 C Yes No

6.OPERATING PRESSURE: 5 atm (gauge) 11. INSULATION HOUSING:

7. Indoor Outdoor 304 SS 316 SS

86

8.DUCT DIMENSIONS: L : 1.38 m W:0.5 m Other(Specify)

9.AIR FLOW DIRECTION: Upward 12. INSULATION THICKNESS: 89mm

10. air velocity: 90 m/min 13. HEATER DIMENSION (mm):

Heater Specifications A: 267 B: 508 C: 708 D: 267 E: 88.9

1. RATING: Volts 240V Phase 3 Kilowatts 18.4 F: 222.25 G: 304.8 H: 76 I: 152.4

2. HEATING ELEMENTS: Tubular(std)

3.HEATING ELEMENT SHEATH MATERIAL: 14. OTHER SPECIAL FEATURES:

Tubular: Corrosion Resistant

INCOLOY (STD) 304 Stainless Steel 15. PRESSURE DROP:

316 Stainless Steel 0.00135 kPa

Other (specify) Nichrome wire (80%Ni, 20%Cr) 16.MODEL NO.:

4. HEATING ELEMENT WATT DENSITY: *DE 101

20W/sq.cm. 22W/sq.cm. 30W/sq.cm.

Other(Specify)

5.NUMBER OF ELEMENTS: 18 * fictional

Mechanical Layout

Figure 3.4.6: Diagram for the duct (left) and schematic diagram of electric heater (right).

c. Mixer

87

The equipment specifications for the mixer were also solved in the energy balance

section. A summary of the specs are given in the table below and a mechanical layout of

the mixer follows.

Table 3.4.4: Equipment Specification Sheet for MixerEQUIPMENT SPECIFICATION SHEET

Item Name Mixing tankQuantity 1Project Description Industrial scaleFunction Mixing vesselMode of Operation Batch

TANK PARAMETERSOrientation verticalShell _______cylindrical______

Shell length, tangent-to-tangent, m ___________2____________Diameter, m 4.40

HeadsTop (i.e. Dished, 2:1 Elliptical)Bottom

________Dished__________________Dished__________

Design Temperature, °C 48 [2]

Design Pressure, atm 1pH range 13.7 [3]

MATERIALS OF CONSTRUCTIONTank Stainless Steel 316Impeller Stainless Steel 316

INSTALLATIONSAgitation

Impeller Type (i.e. rushton) Gate paddleNumber of impellers 1Impeller Diameter, m 3.01Motor Power, hP 102.25Impeller speed, rpm 30Jacket Type simple, unbaffled

FEED/CONTENTSFlowrate, kg/batch (Solids) 9,348.01Initial Temperature, °C 30Flow, inlet TopFlowrate, kg/batch (Water) 12,583.52Initial Temperature,°C 30Flow, inlet TopFlowrate, kg/batch (Mineral solution ) 1,168.50Initial Temperature, °C 30

88

Flow, inlet TopOUTPUT

Flowrate, kg/batch 23,606.09

Final Temperature, °C 30Flow, outlet Bottom

TANK CAPACITYVolume of tank, completely full, m3 100

Mechanical Layout of the Mixer

Figure 3.4.7: Specifications of the mixer.

Figure 3.4.8: Specifications of the impeller.

d. Fermenter

The most important process in the production of high-value animal feeds is the

fermentation. The fermentation is employed in order to increase the protein content of the

substrate through the use of yeast. The enrichment is done through semi-solid state

fermentation. In this process, the substrate contains 55% water. Two fermentation setups

89

are to be employed for the process. Both setups will have the same amount of substrate

and starter culture.

For the design of the fermenter that is implemented in the study, it is shown in the

figure below:

Figure 3.4.9: Design of a Fermenter

The total volume occupied by the materials is shown below:

Thus, the total volume occupied by the materials needed or the working volume

can be calculated using the density of the mixture equal to 463.34 kg/m3. Considering

two fermenters will be used:

V working=mass ossubstrate

density of substrate

V working=

23,606.09 kg2

463.34kg

m3

V working=25.47 m3

Typically, the working volume will be 70-80% of the total fermenter volume. For

the study, a working volume of 70% is assumed. Thus, the total fermenter volume is

calculated below:

90

V Total=V Working

0.70

Eq. 129

V Total=25.47 m3

0.70

V Total=36.38m3

The headspace volume is thus equal to 10.91 m3.

From the geometric proportions of a standard agitation system (Geankoplis, 1995)

as shown in the table 4.4, the measurements are calculated.

The figure below shows the schematic of the agitation system:

Figure 3.4.10: Schematic Diagram of the Agitation System

In calculating the tank diameter:

V Working=π4

D t2 H Eq. 130

Since H = Dt, thus, the equation becomes:

V Working=π4

D t3

36.38 m3=π4

Dt3

91

Dt=3.59 m

Thus, H = 3.59 m.

For the impeller diameter, Da, a Da/Dt = 1/3 is assumed.

Da

D t

=13

Da=13

Dt

Da=13

(3.59 m)

Da=1.2m

To calculate for W:

WDa

=15

W =15

Da

W =15

(1.2 m)

W =0.24 m

To calculate for Dd:

Dd

Da

=23

Dd=23

Da

Dd=23

(1.2 m )

Dd=0.8 m

To calculate for L:

LDa

=14

L=14

Da

L=14

(1.2 m)

92

L=0.3 m

To calculate for C:

CDt

=13

C=13

Dt

C=13

(3.59 m )

C=1.2 m

To calculate for J:

JDt

= 112

J= 112

Dt

J= 112

(3.59 m)

J=0.3 m

In calculating the total height of tank:

V Total=π4

Dt2 H t

Eq. 131

H t=V Total

π4

Dt2

H t=36.38 m3

π4

(3.59 m)2

H t=3.59 m

For the power requirement of the agitated vessels, the viscosity of the slurry

assumed is 200 Pa·s. The turbine is assumed to be operated at 60 rpm or equal to 1 rev/s

since minimal agitation is needed.

N ℜ' =

Da2 Nρμ

93

N ℜ' =

(1.2 m )2(1 revs )(463.34

kg

m3 )200 Pa∙ s

N ℜ' =3.34 ≈ 3

Using curve 1 (for flat six – blade turbine with disk) in figure 3.4-4, the Np = 17 for

N ℜ' =3. Solving for P using equation (3.4-2) and substituting known values, the power

requirement is obtained.

N P=P

ρ N 3 Da5

P=N P ρ N 3 D a5

P=(17)(463.34kgm3 )(0.5

revs )

3

(1.51 m)5

P=7,729.36Js=7.73 kW =10.36 hp

For the material used, the inner cylinder will be stainless steel and the outer

cylinder will be made of glass.

e. Filter Press

Initially, the volume of the filtrate is to be determined. Using the data specifically

the density and the mass of the filtrate, the volume of the filtrate can then be calculated.

The density of the filtrate is taken to be 999.98kg/m3 and the mass of the filtrate cake is

14,285.71 kg.

94

Figure 3.4.11: Schematic Diagram of Filter Press

V filtrate=mas sfiltrate

ρfiltrate

V filtrate=14,285.71 kg×( 1999.98 kg

m3 )V filtrate=14.29 m3

Then the ratio of the mass of the dry cake and the volume of the filtrate is

determined.

C s=14,285.71 (0.65 )

14.29

C s=649.80kgsolid

m3 filtrate

The specific cake resistance is then calculated:

α=4.37 ×109 (−∆ p )0.3

α=4.37 ×109 (101.325 )0.3

α=1.387 ×1011

It is required to determine the total effective filtration area for the further

calculations of the design specification of the filter press. Considering that the material to

be used is cast iron, the effective filtration area per chamber can be determined.

95

Assuming the size of the filter plate to be 1 000 mm, the effective filtration area is 1.74

m2.

Figure 3.4.12: Area and Cake Capacity of Various Sizes of Plate and Frame Filters\

It is also assumed that there will be 30 plates, then the total effective area of the filter press is:

A=effective filtration area× number of plates

A=1.74 m2× 30

A=52.2m2

Calculating for the constant, KP

K P=μα C s

A2(−∆ P)

Assuming that the viscosity of the filtrate is 805.75 x 10-6 Pa.s,

K P=(805.75 ×10−6 ) ( 1.387 ×1011) (649.80)

¿¿

K P=263.03s

m6

96

For constant pressure filtration, the filtration time required is calculated using the equation:

t f =K P

2(V 2 )+BV

It is assumed that the filtration time is approximately 8 hours. Thus the value for B can be calculated:

B=t f−

K P

2(V 2 )

V

B=(8hrs×

3600 s1 hr )−263.03

2(14.29 m3 )2

14.29 m3

B=136.05

Calculating for the filtration rate:

dVdt

=14

¿

dVdt

=14 ( 1

(263.03+14.29 )+136.05 )dVdt

=6.05 ×10−4 m3

s

f. Storage Tank

Storage tanks are necessary in order to store the raw materials prior to mixing in

the mixing tank. Three storage tanks are requires, one for each agro-industrial waste. For

the volume of the agro-industrial wastes:

Cassava Peels

V cassava peels=4,089.76 kgCP×m3

862.5 kg

V cassava peels=4.74 m3 CP

Pineapple Pomace

97

V pineapple pomace=4,089.76 kg PP ×m3

150 kg

V ineapple pomace=27.26 m3 PP

Ipil-ipil leaves

V ipil−ipil leaves=1,168.50 k g IL×m3

265 kg

V ipil−ipil leaves=4.41 m3 IL

For the working volume of each tank, it is assumed that the raw materials occupy 70% of

the tank, thus,

Cassava Peels

V working volume CP ST=4.74 m3

0.7

V working volume CP ST=6.77 m3

Pineapple Pomace

V working volume PP ST=27.26 m3

0.70

V working volume PP ST=38.94 m3

Ipil-ipil Leaves

V working volume ILST=4.41 m3

0.70

V working volume ILST=6.30 m3

It is assumed that the height of the tank is 3 m, thus the surface area of each tank will be:

Cassava Peels

AworkingCP ST=6.8 m3

3 m

AworkingCP ST=2.27 m2

Pineapple Pomace

98

Aworking PP ST=39 m3

3 m

Aworking PP ST=13 m2

Ipil-ipil Leaves

Aworking ILST=6.8 m3

3 m

Aworking ILST=2.27m2

99

CHAPTER FOUR

Economic Analysis

Profitability measures play a crucial role throughout the design process since it

helps the design team select the best design alternatives. This chapter presents the

economic analyses of the feasibility of the plant. In this study, the cost estimation of the

equipment involved in the plant were based from method of Guthrie preliminary design

cost estimations, Lang factors and from credible equipment suppliers in the market.

4.1 Equipment Costing

Expenses of a plant significantly include installation and purchase of equipment.

These values are usually accounted in the total capital investment of the plant and part of

the profitability analysis. Presented in Table 4.1 is the total purchase cost of the major

equipment by the plant. A detailed calculation follows for each equipment follows.

Table 4.1. Total purchase cost of major equipment units.

Equipment Amount (Php)Washer 244,772.00Dryer 409,507.00Eleactric Duct Heater 182,118.72Belt Conveyor 1,732,275.00Hammer Mill 8,336,895.28Storage Tank 8,954,732.74Mixing Tank 1 34,227,354.58Fermenter 53,244,076.37Filter Press 5,426,712.44Pelletizer 5,695,880.61Screw Conveyor 307,075Pumps 793,410Cooling Tower 1,189,945TOTAL 120,744,754.73

Based on literature, the CEPCI index for the base year to be used in this study is

394. On the other hand, the CEPCI used is 580.2 based on 2014 (Chemical Engineering

Plant Cost Index (Cepci), 2015). It is also assumed in the calculations presented in this

paper that $ 1=44.20Php based on January 31, 2015 (XE Currency Converter, 2015).

100

From this, the total permanent investment (fixed capital investment, without the

working capital) and total capital investment of the proposed design can be obtained

using the equation below from (Seider, Seader, & Lewin, 2010).

CTpI=1.05 f LTPI∑

i ( I i

I bi)CPi

CTCI=1.05 f LTCI∑

i ( I i

I bi)CPi

Thus, to obtain the total capital investment, the present cost of each equipment

must be obtained and then summed and multiplied to the appropriate Lang factor and to

1.05. For this study, f LTCI=4.9 as given in the table below.

Therefore, the total capital investment of the project can be calculated if

the total purchase cost of equipment is considered,

CTCI=1.05 (4.9 )(total purchase cost of major equipment)

CTCI=1.05(4.9)¿ 120,744,754.73 ¿CTCI=Php 621,231,763.09

101

4.1.1 Washer

A good equipment in the cleansing of the cassava peels would use a rolling drum

brush washing equipment. The high-pressure sprinkles and the rotating brushes of the

equipment are the ones responsible for the cleaning of the peelings from mud and debris.

However, the required throughput required for the equipment is higher compared to the

raw material input (Equipment Input > 350kg/day) of the plant and the cleaning of the

peels is not completely guaranteed to totally remove the mud on it, thus it is advisable

that manual cleaning of the cassava peels be implored instead. Based from the

Department of Labor and Employment’s (DOLE) current minimum wage matrix for an

agricultural labor classified as category 3 (Bukidnon area: Valenica, Manolo Fortich) the

minimum wage of a worker is set at 284 Php/day for region 10 as shown in the figure

below. This is effective as of January 21, 2015.

Table *. Minimum wage rates for Region 10 with respect to sector and category (Bureauof Labor and Employment Statistics, 2015).

For this particular processing step the total number of workers that the plant

would be employing is 10 individuals. This would total to a labor cost of Php 2,840/day.

This means that for a year the plant will have an annual labor cost, for the washing

process, of Php 1,931,200/yr.

For washing the cassava peelings a total number of 20 workers, each having a

quota of 12.5 kg of cassava peels per day would give a total wage of 5,680 Php/day.

4.1.2 Air Dryer

Drying is employed in order to remove excess moisture content from the raw

materials before fermentation. For the production of high-value animal feeds, air is used

102

as drying medium in a tray dryer. Based on the design specification, the following

specifications were determined:

Table 4.1.2 Specifications for Air Dryer.

Item Name Tray DrierQuantity 1Mode of Operation (i.e. batch) BatchOrientation (i.e. horizontal) VerticalShell (i.e. cylindrical) -Dimensions: height, length, tangent-to-tangent, m

Length = 3.91 m, Height = 4.23 m, Width = 9.03

Diameter, mDesign Temperature, °C >300°CDesign Pressure, atm >10 atmTank (i.e. Stainless Steel 316) Stainless steelJacket (i.e. Stainless Steel 316) -Support (i.e. Saddles, Carbon Steel) -Trays (i.e. Stainless Steel 316) Stainless steelVolume of drier, m3 149.35

The present purchase cost can be calculated.

C2=( 580.2325 )$ 5189 .72C2=$ 9,264.85=Php 409,507

4.1.3 Electric Duct Heater

This section will discuss the cost estimation for an electric heater.

Operating Conditions 5. ELEMENT DIAMETER: 11 mm

1. APPLICATION (Describe in Detail):6. TERMINAL SEALS: Silicone Fluid (260 C)

This air duct heater is used to heat air that will be utilized as the drying fluid for the drying operations

of the cassava peel, pineapple pomace, ipil-ipil leaves and the final animal feed product in the animal feed

production.

7.TERMINAL BOX CONSTRUCTION:

General Purpose

Moisture Resistant

Explosion Resistant

2.AIR FLOW: 36.33 cu.m/min 8. TERMINAL BOX MATERIAL: 316 SS

3.INLET AIR TEMPERATURE: 30 C 9.FLANGE MATERIAL: 304 SS

4.OUTLET AIR TEMPERATURE: 55 C 10. INSULATION HOUSING (Below Flange):

103

5.OPERATING TEMPERATURE: 58 C Yes No

6.OPERATING PRESSURE: 5 atm (gauge) 11. INSULATION HOUSING:

7. Indoor Outdoor 304 SS 316 SS

8.DUCT DIMENSIONS: L : 1.81 m W: 1.33 m Other(Specify)

9.AIR FLOW DIRECTION: Upward 12. INSULATION THICKNESS: 89mm

10. air velocity: 90 m/min 13. HEATER DIMENSION (mm):

Heater Specifications A: 267 B: 508 C: 708 D: 267 E: 88.9

1. RATING: Volts 240V Phase 3 Kilowatts 18.4 F: 222.25 G: 304.8 H: 76 I: 152.4

2. HEATING ELEMENTS: Tubular(std)

3.HEATING ELEMENT SHEATH MATERIAL: 14. OTHER SPECIAL FEATURES:

Tubular: Corrosion Resistant

INCOLOY (STD) 304 Stainless Steel 15. PRESSURE DROP:

316 Stainless Steel 0.00135 kPa

Other (specify) Nichrome wire (80%Ni, 20%Cr) 16.MODEL NO.:

4. HEATING ELEMENT WATT DENSITY: *DE 101

20W/sq.cm. 22W/sq.cm. 30W/sq.cm.

5.NUMBER OF ELEMENTS: 18 * fictional

In order to calculate for the total capital investment of the electric duct heater

using the Lang Method the following factors were used: an MS equipment cost index was

used as the reference since the CE indices is limited to heat exchangers. Second, the

purchase cost of the electric duct heater is calculated using the six-tenths rule given

below:

CP=CP1(Capacit y2

Capacit y1)

0.6

Thus,

104

CP=$1092( 128390 )

0.6

=$ 5378

For the calculation of the purchase cost of the duct, Figure B.2 of Timmerhaus

was used to estimate the fabrication of an aluminum clad duct. This is done by plotting

the cross sectional area of the duct and intersecting it to the line of the material it is made.

The resulting purchase cost, $300/m, is multiplied to the length of the duct given in the

table below as H.

Figure 4: Cost of ductwork as a function of the cross-sectional area, including hangers

and supports.

Note: There is a slight difference with the result for this figure and the one used based on

the book. (Figure presented is from the 4th edition while the one used is from the fifth

edition)

In order to calculate for the CTCI , the equation below was used:

105

CTPI=1.05 f TPI∑ II base

CP

For this particular equipment a Lang factor, f TCI, of 5.93 was used since the

equipment utilizes air, a fluid. The base index used is that of 2011, as stated in the table

below, and an MS cost index for 2014 = 1559.55.

Table *. Summary of the equipment design and its corresponding purchase cost.

EquipmentName

Capacity Material Design Temperature

(°C)

Design Pressure

(atm)

Purchase Cost, $ (MS Index =

1519.6)

Electric DuctHeater

1283 f t 3/min Incoloy 58 5 2,533

Duct 2.42m2 (LW )H=5 m

Aluminum - 5 1500

Total 4,033

Calculating the present cost,

C2=( 1559.551519.6 ) ( $ 4 , 033 )

C2=$ 7,058.82=Php 182, 1194.1.4 Grinder

From energy balance, the power required for each of the three crushers is

approximately 12 kW. According to Seider, for a hammer mill, the following equation is

used for the purchase cost,

CP=3,000W 0.78

where W =feed rate ,ton /hr.

Also, from material balance, feed rate of 2 kg/s maximum was set equivalent to

7.2 tons/hr. Thus,

106

CP=3,000 (305.73 )0.78

CP=$260,426.36

Present cost of the equipment,

C2=( 580.2394 )($ 260,426.36)

C2=$ 20 ,602 . 61=Php 11,463,968.31

4.1.5 Storage tank

The total purchase cost for the calculation the storage tanks is given by the

equation below:

C p=C S0.6

where C is the cost of a single storage tank and S, the surface area of the tank. Since there

are three storage tanks needed the equation above becomes a summation of the three

values of the storage tanks needed.

Thus, C p=Php 105 ,648 (12.530.6+69.890.6+11.300.6 )=Php2,275,935.78

4.1.6 Mixing Tank 1

For this part, the cost of the mixer which uses a gate or anchor paddle will be

evaluated. The specifications of the mixer are given below.

Table 4.1.6 Specifications for Mixing Tank 1.Item Name Mixing tankQuantity 1Function Mixing vesselOrientation verticalShell Cylindrical

Shell length, tangent-to-tangent, m 2

107

Diameter, m 1.3656Tank Stainless Steel 316Impeller Stainless Steel 316

Impeller Type (i.e. rushton) Gate paddleNumber of impellers 1Impeller Diameter, m 0.9104Motor Power, hP 6.54Impeller speed, rpm 90Jacket Type simple, unbaffled

Volume of tank, completely full, m3 2

In addition, A=5.86 m2=63.06 f t 2 and P=6.54 hp=5 kW .

The purchase cost of the equipment can be determined by adopting Equation 6.7

of Chemical Engineering Design by Sinott.

Ce=C Sn

where Ce=purhcased equipment cost ,

S=characteristic ¿¿

C=cost constant

n=index for thetype of equipment

From Table 6.2 (Sinott, 2005), using the parameters for an agitator since those for

a mixer is not available.

108

For the propeller,

Ce=($1900 ) (5 )0.5

Ce=$ 4,248.53

For the turbine,

Ce=($3000 ) (5 )0.5

Ce=$6,708.20

For the vessel, the calculation of the purchase cost is done by obtaining the bare

module cost from figure 2, and the material and pressure factor from figure 3. This can be

summarized by the equation below:

Purchase cos t=(bare cost ¿ figure ) ( Material fatcor )( pressure factor)

Figure 2. Equipment costing for vessels.

For vessels with diameter approximately 2m, curve 2 is used. Thus,

euipment cost=$ 8,000, since vessel height is 1.565 m approximately 2 m.

109

Figure 3. Material and Pressure factors.

where MF=2.0 and PF=1.0 Substituting the obtained values,

Ce=$ 8,000 (2.0 )(1.0)

Ce=$16,000

Summing up the calculated purchase cost,

Ce=$ 4,824.53+$6,708.2+$ 16,000

Ce=$27,532.73

Calculating the present cost of the equipment with base cost index 2004 =444.2.

Thus,

C2=( 580.2444.2 )($ 27,532.30)

C2=$ 35,961.82=Php 1,589,513

This mixing tank is adjusted to accommodate a bigger production output from its initial capacity of 600L the mixer has been upgraded to a capacity of 100, 000L. Using the six-tenths rule to calculate the new purchase cost of the mixer yields the following new purchase cost.

Cp=( 100000600 )

0.6

∗Php1 ,589 ,513=Php 34 , 227 , 936

4.1.7 Fermenter

For the design of the fermenter that is implemented in the study, it is shown in the

figure below:

110

In the fermenter design, a jacketed, agitated stainless steel vessel is to be used and

the reactor volume is said to be around 2.426921 m3 or 641.125 gallons. From the

description of the fermenter design, it can be noted that they have the same description

with an autoclave. An autoclave is predominantly a vertical, cylindrical stirred – tank

reactor, which can be jacketed as a means of transferring heat to or from the vessel.

Hence, the purchased cost of an autoclave is obtained. For a stainless steel autoclave

having turbine agitator and heat transfer jacket, the purchase cost equation is shown

below (Seider):

CP=1,980 S0.58

where:

CP = purchased equipment cost on a CE Index = 500

S = size parameter in gallons

Thus, the purchased cost, Cp, based on CE Index = 500 is calculated.

CP=1,980 S0.58

CP=1,980(641.12)0.58

C p=$84,080

111

Calculating present cost of equipment, the CEPCI for 2014, which is 580.2, is

used against the CEPCI = 500, in which the data is based.

C2=CP( Ii

I bi)

C2=( 580.2500 )($ 84,080)

C2=$ 97,566.43=Php 4,312,436

Using the same process as the mixer the up scaled cost of the fermenter from its initial

capacity of 2 cubic meters to a 131.91 cubic meter fermenter is equal to 52, 244, 076Php

4.1.8 Filter Press

Based from the calculations conducted for the design specification of a filter

press, it was calculated that the total effective filtration area is 26.1 m2 or 280.938 ft2. The

purchase cost for the filter press can be calculated using the equation taken from Product

and Process Design Principles by Seider:

Ce=C Sn

The cost constant, C and the index, n is obtained from Chemical Engineering

Design by R.K. Sinnott and are given below:

112

Substituting values:

Ce=8800(26.1)0.6

Ce=$62,296.98

Calculating for the present cost:

C2=( 580.2444.2 ) ($62,296.98 )

C2=$ 81,370.35=Php 3,580,295

Up scaled version of the filter press was made from a filtration area capacity of 26.1 square meters to a 52.2 square meters capacity, and the calculated cost for this is equivalent to 5,426,712.44Php.

4.1.9 Pelletizer (Extruder)

A specification for the pelletizer or extruder is given by Table4.1.10. Detailed

calculation follows.

Table 4.1.9 Technical specifications for pelletizer.Parameter Value

Production capacity, kg/h for pelleted feed 10,000Power consumption for pelleted feed, kW/h 62Consumption of steam, kg/h 100Length, mm 2400Width, mm 2400Height, mm 2850

113

Using Lang method to solve for the Total Capital Investment for the extruder equipment:

The f.o.b. purchase cost for this type of equipment is at $16 800 from 2011 price.

To calculate the present cost of this equipment, Marshall and Swift index of

1519.6 and 1559.55 for 2011 and 2014 is used. Thus,

C2=( 1559.551519.6 )($126,078.44 )

C2=$ 129,393.02=Php 5,695,880.61

4.1.10 Pumps

In calculating for the cost of the pumps used in the production process, the

following operating conditions are to be considered:

Table 4.1.10.1: Operating Conditions for Pumps

Flow rate (gpm) Head (ft)Pump for cooling water 3.726 50Pump for drain water 5.837 50

114

The cost estimation is initiated by estimating the size factor. Calculating for the

size factor of the pump:

S=Q H 0.5

The flow rate and the pressure head of the fluid are 3.726 gpm and 50 ft for the pumps for

the cooling water. Substituting values:

S=(3.726)(50)0.5

S=26.35

Calculating for the base cost of the centrifugal pump using the equation given in Product

and Process Design Principles by Seider:

CB=exp ¿

Substituting equations:

CB=exp ¿

CPC=$2647.74

Calculating for the purchase cost:

CPC=FT FM CB

FT is the pump type factor while FM is the material factor. These values are given the in

the tables provided in the Product and Process Design Principles by Seider.

115

Calculating for the purchase cost:

CPC=(1.0 ) (1.0 ) (2432.83 )

CPC=$2432.83

For the pumps for drain water, the volumetric flow rate of the water is 5.837 gpm.

The head is taken to be 50 ft. Calculating for the size factor:

S=Q H 0.5

S=(5.837)(50)0.5

116

S=41.274

Calculating for the base cost:

CB=exp ¿

CB=$2378.43

Then, calculating for the purchase cost:

CPC=FT FM CB

CPC=(1.0 ) (1.0 ) (2378.43 )

CPC=$2378.43

It is noted that 4 pumps are used in the entire production process. The thirds pump

is considered to have same parameters as the pump for the drain water. The last pump to

be used is a slurry pump. For the slurry pump, the following information will be used:

Table 4.1.10.2: Cost for Slurry Pumps

Model SRL-C Slurry PumpSupplier Allis-ChalmersSupplier Address Ontario, CanadaPrice $5,000

Calculating for the total purchase cost for the pump:

T PC=2432.83+2378.43+2378.43+5000

T PC=$ 12,189.72

Thus, present purchase cost of the pumps is,

C2=( 580.2394 ) ($ 12,189.72 )

C2=$ 17,950.45=Php 793,410

117

4.1.11 Belt Conveyor

The purchase cost for belt conveyors is calculated based on the equation given by

Seider on Table 16.32 (Seider, Seader, & Lewin, 2010),

CP=16.9WL

where W =width ,∈.

L=length , ft

Specifications are listed in Table 4.1.11 below.

Table 4.1.11 Specifications for belt conveyor. Specification DataLength, L, ft 32.81Width, W, in. 24Speed, m/s 0.508Capacity, m3/batch 0.822

Substitute known values to equation above,

CP=16.9 (24∈. ) (32.81 ft )

CP=$13,307.09

For its present purchase cost,

C2=( 580.2394 ) ($ 13,307.09 )

C2=$ 19,595.87=Php 866,137.55

There are two belt conveyors proposed in the design, however, it is assumed that

the two have similar cost. Thus, for the belt conveyors,

C2=Php 866,137.55(2)

C2=Php1,732,275

118

4.1.12 Screw Conveyor

Screw conveyors are suitable for transport of sticky and abrasive solids. Thus, for

the transport of the milled raw materials it is used. Based on heuristics, the equipment is

limited to 3.81 m or so because of shaft torque strength (Rules of Thumb: Summary),

therefore its length would be L=3.81m=12.5 ft.

From Product and Process Design Principles by Seider (Table 16.32), the

purchase cost of a screw conveyor can be calculated by the following equation,

CP=55.6D L0.50

where D=diameter ,∈.

L=length , ft

Based on heuristics, a 304.8 mm (12 in.) diameter conveyor can handle 283–8495

m3/h 1000–3000 ft3/h, at speeds ranging from 40 to 60 rpm (Rules of Thumb: Summary).

For this process, D=12∈. and N=50 rpm. Substituting known values to equation above

gives,

CP=55.6 (12∈. ) (12.5 ft )0.50

CP=$2,358.91

Thus, the present purchase cost is,

C2=( 580.2394 ) ($ 2,358.91 )

C2=$ 3,473.70=Php 153,537.74

In the proposed deisgn, two screw conveyors are used. Thus,

C2=Php153,537.74 (2)

C2=Php307,075

119

4.1.13 Cooling Tower

In the process, a cooling tower is to be used as a storage of cooling water to be

used during fermentation. For one batch of production, the cooling water needed is

30.463 m3 or 8,047.47 gal. The design of the cooling tower is based on heuristics. For

storage tanks having a capacity of 1000-10000m3 , a horizontal tank on concete support is

used. From Seider's Product and Process Design Principles, for a carbon steel cone roof

storage tanks operating up to 3psig and having a range, S, of 10,000 to 1,000,000 gallons,

the solution for the purchased cost is given by:

Cp = (210)(V)°.51

Cp = (210)(8047.47)°.51

Cp = $20,611.30

Calculating present purchase cost of equipment,

C2=( 580.2444.2 )($ 20,611.30)

C2=$ 26,921.83=Php 1,189,945

4.2 Total Capital Investment Estimation

The total capital investment of a chemical plant is analogous to the purchase price

of a new house where the price includes purchase of land, building, permit fees,

excavation of the land, improvements to the land to provide utilities and access, etc. It is

a one-time expense for the design, construction, and start-up of a new plant or a

renovation of an existing plant (Seider, Seader, & Lewin, 2010).

Using Lang factors to account the cost of installation, construction, etc, the total

capital investment can be obtained. Lang factor values are obtained from Sinnott (2005).

The total purchase cost of the major equipment is summarized in Table 4.2. The fixed

capital is the sum of direct and indirect costs. The Lang factors for a fluid- solid type

plant is shown and is used to calculate for the direct costs (total physical plant cost) used.

120

Table 4.2. Lang factors used for estimation of project fixed capital cost (Sinott, 2005). Item Fluids-Solids Cost (Php)

1. Major Equipment, Total Purchase Cost PCE 16,314,848f1 Equipment erection 0.45 7,341,682f2 Piping 0.45 7,341,682f3 Instrumentation 0.15 2,447,227f4 Electrical 0.10 1,631,485f5 Buildings, process 0.10 1,631,485f6 Utilities 0.45 7,341,682f7 Storages 0.20 3,262,970f8 Site development 0.05 815,742f9 Ancillary Buildings 0.20 3,262,970

2. Total physical plant cost (PPC) = PCE(1+f1+…+f9) 3.14 51,391,771f10 Design and Engineering 0.25 12,847,943f11 Contractor’s Fee 0.05 2,569,589f12 Contingencies 0.10 5,139,177

3. Fixed Capital = PPC(1+f10+f11+f12 ) 1.40 71,948,4804. Working Capital = FC(0.1) 7,194,8485. Total Capital Investment = FC+WC 79,143,328

4.3 Profitability

The feasibility of the project can be determined by subjecting it to profitabitly

analysis. Here, a discounted profitability analysis will be done and the discounted

payback period and the discounted cumulative cash position will be obtained. After

obtaining the required total capital investment for the project, the total product cost will

now then be obtained. Total product cost will be divided into the following: direct

production cost, fixed charges, plant overhead costs, administrative expenses, distribution

and marketing expenses, research and development, financing and gross earnings

expense.

Direct Production Costs

a. Cost for raw materials

Table **. Annual cost for raw materials

Raw Material UnitAmoun

t per batch

Amount per year, (110

batches/year)

Cost in Php/unit

Actual cost in Php

Cassava peels kg 5,684.35 625,278.50 0.76 477,087.50

121

Pineapple pomace kg7,431.21 817,433.10 3.33

2,720,090.38

Ipil-ipil leaves kg 1,240.63 136,469.30 0.50 68,234.65

S.cerevisiaea kg 236.06 25,966.60 95.002,466,827.0

0Ammonium sulphateb kg 35.06 3,856.60 6.80 26,224.88Magnesium sulphateb kg 10.52 1,157.20 17.56 20,320.43

Manganese sulphatec kg 1.17 128.70 28.32 3,644.78Potassium phosphatec kg 18.70 2,057.00 44.68 91,906.76TOTAL 5,874,336.3

9aPrices are obtained from sigmaaldrich.com. bPrices are obtained from icis.com.cPrices are obtained from ychsenfa.en.alibaba.com.

b. Operating Labor

8 process are considered in the calculation for the total operating labor cost. Given

in the table below, 66 operators are required for the operations of these processes. In the

calculation for the cost, a daily salary of Php 284.00 is considered based from the

minimum rate of Region X effective last January 1, 2015. It is assumed that the plant will

operate for 24 hours meaning it will have 2 shifts (day and night) and for 300 days in a

year. The given table below shows the summary of the operating labor cost of each

process.

Table **. Information on operating labor of the plant.Unit Process Equipment No. of

operators/shift/day

No. of operators/da

y

Total daily

salary of operators,

Php

Total annual

salary of operators,

PhpWashing None 1 20 5680 1,874,400Drying Air Dryer +

Electric Duct Heater

2 8 2272 749,760

Grinding Hammer Mill

2 4 1136 374,880

Mixing Mixer 2 6 1704 562,320Fermentation Fermentor

+ starter 3 6 1704 562,320

122

culture tanks +

Filtration Plate and Frame Filter Press

2 12 1704 562,320

Pelleting Extruder 2 4 1136 374,880Drying of Animal Feed Pellets

Air Dryer 2 6 1704 562,320

Total Cost of Operating Labor 5,623,200

c. Direct supervisory and clerical labor

Supervisory labor includes the managers and supervisors assigned on different

processes. Clerical labor includes workers in the administration. Operating cost on

supervisory and on clerical labor is taken to be 25% of the total cost of operating labor

(Timmerhaus, 2003).

Supervision∧clerical labor cost=0.25(Operating labor cost )

Supervision∧clerical labor cost=0.25(Php 5,623,200)

Supervision∧clerical labor cost=Php 1,405,800

d. Laboratory Charges

Laboratory charges is taken to be 10% of the total operating labor cost

(Timmerhaus, 2003).

Laboratory charges=0.10 (Php 5,623,200 )

Laboratory charges=Php 562,320

e. Maintenance and repair

Cost of maintenance and repair of machineries in the plant are taken to be 2% of

the fixed capital investment (Timmerhaus, 2003).

Cost of maintenance∧repair=0.02 FCI

123

¿0.02(Php 71,948,480)

¿ Php 1,438,970

f. Power and utilities

The main utilities used in this plant are electricity, steam and water. The amount

of electricity, steam and water for the yearly operation are obtained from the equipment

specification and material balance calculations. For 2015 the current rate for power

consumption of a high load industrial plant is at 0.1153Php/kWh + 2623.43 Php/mo

(ERC, 2015). Total power consumption of plant is seen at 236.93 kW not including

lighting and other devices.

Calculating for the annual cost of electric consumption

cos t electric=236.93 kW (16hoursday )( 0.1153 Php

kWh )(330daysyr )+ 2,633.43 Php

mo (12moyr )=175,720 Php / yr

Therefore, costs for the power consumption for a year would reach to 175,720 Php/yr and direct production can then be calculated.

Direct productioncost=Cost of raw materials+operating labor+Supervisory labor+Laboratory charges+Maintenance∧repair+Power∧utilities

Substituting values, the direct production cost is equal to:

Direct productioncost=Php12,084,955

Fixed Charges

a. Depreciation

Using straight-line method, it is assumed that the salvage value of the plant will

be zero since there will be no anticipated value by the end of its useful life and its life is

estimated to be 10 years. Thus, annual depreciation cost can be calculated using the

following equation,

124

Depreciation , D=FC−Sn

where: FC=¿capital

S=salvage value

n=life of the project

Substitute known values,

Depreciation , D=Php 71 ,948 , 480−010

¿ Php 7 ,194 ,848

b. Local taxes

This may be estimated as 3% of the fixed capital investment (Timmerhaus, 2003).

Local taxes=0.03 FCI

Local taxes=0.03 (Php71 , 948 ,480)

Local taxes=Php Php2,158,454

c. Insurance

This may be estimated as 1% of the fixed capital investment (Timmerhaus, 2003).

Insurance=0.01 FCI

Insurance=0.01(Php 66,700,819.58)

Insurance=Php719,485

Therefore, total fixed charges can be calculated

¿charges=Depreciation+Local taxes+ Insurance

¿7,194,848+2,158,454+719,485

¿charges=Php 10,072,787

125

Plant Overhead Costs (POC)

This cost is estimated to be 50-70% of the sum of operating labor, supervision and maintenance (Timmerhaus, 2003). Taking the value as 60% of the operating labor, supervision and maintenance,

Plant Overhead Cost=0.50(Operating labor+Supervisory cost+Maintenance)

¿0.50(5,623,200+1,405,800+1,438,970)

POC=Php 4,233,985

Administrative Expenses

This cost is estimated to be 15% of the sum of operating labor, supervision and maintenance (Timmerhaus, 2003).

Administrative Expenses=0.15 (Operatingl abor+Supervisory cost+Maintenance)

Administrative Expenses=0.15 (5,623,200+1,405,800+1,438,970)

Administrative Expenses=Php 1,270,196

Distribution and Marketing Expenses

This cost is estimated to be 2-20% of the total product cost (Timmerhaus, 2003).

Total DistributionCost=0.02 TPC

Total DistributionCost=0.02 (31,111,305.69 )

Total DistributionCost=Php 622,226.11

Research and Development

This cost is estimated to be 5% of the total product cost (Timmerhaus, 2003).

Total Research∧Development Cost=0.05 TPC

Total Research∧Development Cost=0.05 (31,111,305.69 )

126

Total Research∧Development Cost=Php 1,555,565.29

Financing

This cost is estimated to be 0-10% of the total capital investment (Timmerhaus,

2003). As of December 11, 2014 in the Philippines, the interest rate on borrowed money

is 4% for the year 2015 as set by the Bangko Sentral ng Pilipinas (BSP) (Taborda, 2014).

Financing Cost=0.04(TCI )

Financing Cost=0.04 ( Php79,143,328 )

Financing Cost=Php 3,165,733

Gross-earnings Expense

In the Philippines, the gross earning expense is 30% of the taxable income.

Moreover, the taxable income is calculated by obtaining the difference between the total

income and the total product cost.

Calculation of the total income

This section shows the total income or revenue generated from the production of

biodiesel.

Table . Calculation of the total income/total revenue.Product Annual production (kg) Price (Php/kg) Income (Php)

HVAF 228900 30 5,722,500

TOTAL INCOME Php 5,722,500.00

Calculation of total product cost

This section shows the calculation of the total product cost. All components are

equated and arranged so that the total product cost can be determined. Moreover, the total

127

product cost per kg of HVAF is calculated. This serves as basis for the market price of

HVAF from agro-industrial wastes.

Component Cost

Direct Production Costs 12,084,955

Fixed Charges 10,072,787

Plant Overhead Cost 4,233,985

Administrative Expenses 1,270,196

Distribution and Marketing Expenses 0.02(TPC)

Research and Development 0.05(TPC)

Financing 3,165,733

Total Product Cost TPC

Total Revenue 5,722,500

Gross-earnings expense 0.30(2,100,000 - TPC)

TPC=12,084,955+10,071,787+4,233,985+1,270,196+0.02 TPC+0.05 TPC+3,165,733+5,722,500+0.3 (5,722,500−TPC )

TPC=Php31,111,306

Total product cost for HVAF= TPCTotal Amount of HVAF production

Total product cost for HVAF=31,111,306228900

Total product cost for HVAF=Php 135.92

Calculation of the net present worth and discounted payback period

The criteria for profitability will be the net present worth and the discounted

payback period. The following assumptions are considered in the calculations:

a. the plant will operate at 80% capacity during the first year

b. the plant will operate at 90% capacity during the second year

c. the plant will operate at 100% capacity in the succeeding years

128

d. the time-value of money is 4% per year (Banko Sentral ng Pilipinas, 2015)

I. Net Present Value

Calculating for the net cash flow in first and second year and the succeeding

years:

net cash flow first year=0.8 (5,722,500 )−31,111,306=−Php 26,247,181

net cash flow second year=0.9 (5,722,500 )−31,111,306=−Php25,961,056

net cash flow for the succeeding years=(5,722,500 )−31,111,306=−Php 25,388,806

Present value of all positive cash flows=¿

Since the values for the net cash flow are lesser than zero, then the project is not

profitable.

II. Discounted Payback Period

The following are previously calculated:

fixed capital investment = Php 71, 948, 480

net cash flow in the first year = Php 26,247,181

net cash flow in the second year = Php 25,961,056

net cash flow in the succeeding years = Php 25, 388, 806

These values are not “given” interest yet and these values gain interest through

time at a nominal interest rate of 4% per year. Assuming that these values are periodic

and that the plant will deposit money in the bank every month, then an effective interest

rate, i’ may be calculated as follows:

(1+i ' )12=1.04

i'=0.0033

129

This means that the nominal interest of 4% per year is equal to 0.33%

compounded monthly. During the first year of operation, the plant will be able to deposit

a monthly value of: 26,247,181

monthly deposit for 1 st year=Php 26 , 247 ,18112

=Php 2 ,187 ,265

Therefore, the remaining time in order to recover the remaining balance of the

investment is

71 , 948 , 480=2, 187 , 265 ×1.0033n−1

0.0033

n=31 months

Thus, the discounted payback period is 31 months or 2 years and 7 months. This

is the time required to attain breakeven. The desired payback period for engineering is

less than 2 years. Thus, the calculated value implies that it is not desired because the

calculated value exceeds the desire payback period.

130

CHAPTER FIVE

Environmental Consideration

This section will contain a 3-4 page summary of the LCA or any waste treatment

design outputs that your group had accomplished as requirement in ChE 52 (Ind. Waste

Management and Control).

5.1. Definition of Goal and Scope

The processes designed in the feasibility study on production of high value animal

feeds aim to address the local community’s alternative source of animal feed that is

cheaper yet packed with the right amount of nutrients for their pigs. The industry uses

biological resources, S. cereviseae, and agro-industrial wastes, cassava peels and

pineapple pomace, as raw materials for the production of high value animal feeds. This

study is intended to assess whether the process designed for the production of high value

animal feeds from agro-industrial wastes is economical and safe both to the surrounding

areas and humans.

5.2. Life Cycle Inventory Analysis

5.3. Life Cycle Impact Assessment

5.3.1. Characterization of Global Warming Impact per Sub-process

5.3.2 Grouping of Indicators

5.4. Life Cycle Interpretation

5.4.1 Evaluation of the Completeness, Sensitivity and Consistency of Data

5.4.2 Findings

131

Wastewater Treatment Design

Total wastewater ¿ plant=43.04m3

day=4.986 x10−4 m3

s

Process Flow Diagram

Adapted from: watercorporation.com.au 2013 (Typical Wastewater Treatment Plant Flow Diagram)

Layout

132

Figure*: Wastewater treatment layout.

133

CHAPTER SIX

Conclusions and Recommendations

The principal conclusions of the design study should be presented, together with a

clear statement of recommendations, accompanied by justifications, for management.

Again, you can use your imagination here to connect it with the letter of transmittal.

134

135

Acknowledgments

The proponents of this study would like to extend their gratitude to the following

persons and institutions for the valuable support and completion of this study:

Dr. Hercules R. Cascon, the group’s research adviser and FYPS instructor, for his

guidance and advices in the development and success of this study;

Dr. Maria Rosario Mosqueda and the Agriculture lab Pool Technicians for

allowing use their automatic distillation and titration apparatus.

Mr. and Mrs. Joel Allera, Mr. and Mrs. Estrada, Mr. and Mrs. Jose B.Torayno,

and Mr. and Mrs. Yap, the parents of the proponents, for their undying support and

understanding during the conduct of this research;

Classmates and friends of the proponents, for their physical and moral support;

And above all, to our Almighty God, for the strength and wisdom, and for making

the proponents’ final year project study successful.

136

137

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Curriculum Vitae

VERA LUWESA M. ALLERAZone 2, Ampenican, Salay, Misamis [email protected]

PERSONAL INFORMATIONBirth date: September 20 1993Birth place: Cagayan de Oro CityAge: 21Civil status: SingleReligion: Roman CatholicGender: FemaleNationality: Filipino

Father: Joel B. Allera Occupation: PolicemanMisamis Oriental Police Provincial Office, San Martin, Villanueva, Misamis Oriental

Mother: Mary Ann M. Allera Occupation: TeacherDepEd Region X, Mastersons Ave., Cagayan de Oro City

EDUCATIONAL BACKGROUNDTertiary Bachelor of Science in Chemical Engineering

Xavier University – Ateneo de CagayanCorrales St., Cagayan de Oro CityS.Y. 2010-2015

Secondary Salay National High SchoolSalay, Misamis Oriental2007-2010St. John the Baptist High SchoolLagonglong, Misamis OrientalHigh School2006-2007

Elementary (Intermediate) Salay Central SchoolSalay, Misamis Oriental2005-2006, 2003-2004East City Central School

148

Lapasan, Cagayan de Oro CityElementary2004-2005

Elementary (Primary) Salay Central SchoolSalay, Misamis Oriental2001-2003East City Central School,Lapasan, Cagayan de Oro City2000-2001

RELATED EXPERIENCESPOSITION COMPANY NAME COMPANY

ADDRESSINCLUSIVE

DATES

On-the-Job-Trainee

First Industrial Plastics Venture, Incorporated (FIPVI)

Alwana Business Park, Cugman, Cagayan de Oro City

April 11–27, 2013

On-the-Job-Trainee

DENR-Environmental Management Bureau Region 10

DENR 10 Compound, Macabalan, Cagayan de Oro City

April 7-June 2, 2014

AWARDSNAME/TYPE OF

AWARDAWARD-GIVING BODY DATE

AWARDED KSSR Award Kinaadman Research Center January 2015

Fr. Araneta Scholar Office of Scholarship and Finance Aid

S.Y. 2014-2015

Academic Scholar Office of Scholarship and Finance Aid

S.Y. 2010-2013

ORGANIZATIONSPOSITION NAME OF ORGANIZATION INCLUSIVE

DATES

Member Junior Philippine Institute of Chemical Engineers

S.Y. 2010-2015

Member Christian Life Community S.Y. 2013-2015

SEMINARS/TRAININGS ATTENDEDDATE

Capillary Electrophoresis for the Analysis of Bioactive CompoundsAVR 3, SBM Building, Xavier University – Ateneo de

January 21, 2015

149

Cagayan, Cagayan de Oro City

Lecture on EntrepreneurshipLittle Theater, Xavier University – Ateneo de Cagayan, Cagayan de Oro City

November 21, 2014

Earth Day Environmental SymposiumDENR-10, Macabalan, Cagayan de Oro City

April 22, 2014

SIGMA: The Engineering Youth’s Response to the Nation’s Call for a Greener NationActivity Center, Ayala Centrio Mall, Cagayan de Oro City

February 22, 2014

PANAGDUYOG Chemical Engineering SummitMalasag Gardens, Ecovillage, Cugman, Cagayan de Oro City

January 31-February 2, 2014

SKILLS Knowledgeable in basic computer

software (MS Office) Knowledgeable in Visual C++

programming Knowledgeable in DraftSight and has a

background on AutoCAD for designing purposes

Knowledgeable in MS Visio for process and plant designs

Knowledgeable in ChemSep and VenSim for distillation and reaction simulations

Able to solve chemical engineering problems (unit operations, unit processes)

Knowledgeable on process equipment design and plant design

Has a background on industrial waste and management control

Has a background on safety management/engineering, environmental engineering, food safety engineering, solid waste management, and wastewater treatment

Willing to be trained Good communication and interpersonal

skills Can work under pressure and can

handle time effectively Able to work effectively alone or in a

team Can do research and development tasks

CHARACTER REFERENCEAvailable upon request.

150

MARY ROSE M. ESTRADAZone 4 Bayabas Boundary, Cagayan de Oro [email protected]+639268084682

EDUCATIONAL ATTAINMENTBachelor of Science in Chemical EngineeringXavier University – Ateneo de CagayanCorrales Avenue, Cagayan de Oro CityS.Y. 2014-2015

SKILLS Good communication skills in oral and in

written Good interpersonal skills Ability to facilitate and lead a group Knowledgeable in making risks,

environmental and cleaner production assessment reports

Knowledgeable in basic chemical engineering plant design, process equipment design and control system

Knowledgeable in wastewater engineering Good public speaking skills Ability to analyzed experimental data Ability to work under less supervision

Good background in environmental engineering

Knowledgeable in solid waste management Ability of conducting laboratory works for

quality control Skills in identifying, formulating and

solving chemical engineering problems Skills in Basic Computer Drafting

(AutoCAD, Vensim, ChemSep) Knowledgeable in basic computer software

(MS Word, MS Excel, MS Powerpoint) Skills in managing events Skills in organizing events Ability to work under pressure Able to adapt to various work conditions

RELATED EXPERIENCES

POSITION COMPANY NAMECOMPANY ADDRESS

INCLUSIVE DATES

OJTSHEMBERG Marketing Corporation

Mandaue City, Cebu April 21 – May 29, 2014

OJTFirst Industrial Plastic Ventures Inc.

Cugman, Cagayan de Oro City

April 11 – 27, 2013

AWARDSNAME/TYPE OF AWARD AWARD-GIVING BODY DATE AWARDED

151

KSSR Award XU - Kinaadman Research Center December 17, 2014 Scholar XU – Office of the Scholarships and

Financial AidA.Y. 2010 – 2011 to A.Y. 2012 - 2013

ORGANIZATIONSPOSITION NAME OF ORGANIZATION INCLUSIVE DATES

Internal Vice PresidentJunior Philippine Institute of Chemical Engineers

A.Y. 2014-2015

PresidentJunior Philippine Institute of Chemical Engineers

A.Y. 2013-2014

Board of DirectorAssociation of the College of Engineering Students

A.Y. 2013-2014

Board MemberPhilippine Institute of Chemical Engineers -Junior Chapter Mindanao

A.Y. 2013-2014

Student Representative Philippine Institute of Chemical Engineers A.Y. 2013-2014

152

JOHDEM I. TORAYNOZone 7, Patag, Cagayan de [email protected]+639059343099

EDUCATIONAL ATTAINMENTBachelor of Science in Chemical EngineeringXavier University – Ateneo de CagayanCorrales Avenue, Cagayan de Oro CityS.Y. 2014 - 2015

SKILLS Knowledgeable in Process and Plant

Design Knowledgeable in material balancing

Knowledgeable in Vensim, ChemSep and CAD software

Fluent in English and basic Nihong-go

Good background in MS applications Knowledgeable in different plant processes and unit operations

Good communication skills Organized Capable in laboratory works and analysis

for Quality Control Knowledgeable in herbal medicine and

basic pharmaceutical terms Capable in working in teams or alone Highly adaptable to social environment Able to read basic Kanji characters Optimistic Resourceful Capable in video editing Dedicated to assigned task Capable in reading maps Creative Critical thinker Capable in facilitating people

RELATED EXPERIENCE/SPOSITION COMPANY NAME COMPANY ADDRESS INCLUSIVE DATES

Trainee Nestle Philippines Inc. – CDO Factory

Tablon, Cagayan de Oro City April 7 – June 5, 2014

Trainee First Industrial Plastic Ventures Incorporated

(FIPVI)

Gusa, Cagayan de Oro City April 2013

AWARDSNAME/TYPE OF AWARD AWARD GIVING BODY DATE AWARDEDKinaadman Research Grant Kinaadman Research Center January 2015

Pryce Grant (Scholarship) Xavier University Alumni Office November 2014Top Book Borrower Xavier University Libraries November 19, 2014

DOST RA7687 S&T Scholarship DOST R-10 Office April 2009ORGANIZATIONS

POSITION NAME OF ORGANIZATION INCLUSIVE DATESRepresentative (Youth with Oro Youth Development Council March 2014 - present

153

Special Needs Sector) (OYDC)

Treasurer Circulo de Arte AY 2014 - 2015

Student Program Officer (SPO) Kristohanong Katilingban sa Pagpakabana – Social Involvement Office(KKP-SIO)

AY 2014 - 2015

DACA HeadDMSA Head

Junior Philippine Institute of Chemical Engineers (JPIChE)

AY 2013 – 2014AY 2012 - 2013

SEMINARS/ TRAININGS ATTENDED DATESLecture on EntreprenuershipLittle Theater, Xavier University, Cagayan de Oro City

November 21, 2014

Youth Advocacy Series: Solid Waste ManagementCity Tourism Hall, Cagayan de Oro City

June 5, 2014

Basic Microbiology Seminar and Hazard Analysis Critical Control Point SeminarNestle Philippines Training Center, Cagayan de Oro City

April 23, 2014

Hazard Identification Risk Assessment and Determining Controls Orientation SeminarNestle Philippines Training Center, Cagayan de Oro City

April 22, 2014

Food Handler’s SeminarCity Health Office, Cagayan de Oro City

April 3, 2014

Sigma: The Engineering Youth ResponseCentrio Activity Center, Cagayan de Oro City

February 22, 2014

Panagduyog (Mindanao-wide Chemical Engineering Summit)Xavier University – Ateneo de Cagayan, Cagayan de Oro

January 30–February 1, 2014

3rd Greeneration Summit: A Gathering for Youth Empowerment on Climate Change for MindanaoCapitol University, Corrales Avenue Osmeña Street, Cagayan de Oro

August 2, 2013

The Best of Me: Personal Development Seminar and WorkshopAVR 4, Xavier University, Cagayan de Oro City

February 13, 2013

References available upon request

154

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