CO-PROCESSING OF JATROPHA OIL WITH LINEAR RUN DIESEL

7/25/2019 CO-PROCESSING OF JATROPHA OIL WITH LINEAR RUN DIESEL

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CO-PROCESSING OF JATROPHA OIL WITH

LINEAR RUN DIESEL

A PROJECT REPORT

Submitted by

YESWIN RAJA.R 212712203086

RASHESH GUPTA.D.V 212712203501

REMO PRATHAP 212712203502

In par t ia l fu l f i l lment for the award of the degree

Of

BACHELOR OF TECHNOLOGY

In

CHEMICAL ENGINEERING

SRI VENKATESWARA COLLEGE OF ENGINEERING

PENNALUR, SRIPERUMBUDUR-602 117

ANNA UNIVERSITY: CHENNAI 600 025

APRIL, 2016




BONAFIDE CERTIFICATE

Cert i f ied that this project report “CO-PROCESSING OF

JATROPHA OIL WITH LINEAR RUN DIESEL” is the

bon afid e wor k of “YESWIN RAJA.R, RASHESH GUPTA.D.V

and REMO PRATHAP” who carried out the project work

under my supervis ion.

SIGNATURE SIGNATURE

Dr.N.Meyyappan, M.E.,Ph.D , G.Hima Bindu, M.Tech.

PROFESSOR & ASSISTANT PROFESSOR

HEAD OF THE DEPARTMENT & SUPERVISOR

Dept of Chemical Engineering Dept of Chemical Engineering

Sri Venkateswara College of Engg Sri Venkateswara College of Engg

Pennalur Pennalur

Sriperumbudur Sriperumbudur

Tamilnadu-602117 Tamilnadu-602117

INTERNAL EXAMINER EXTERNAL EXAMINER

Date: Date:




ACKNOWLEDGEMENT

We express our sincere thanks to Professor S. GANESH

VAIDYANATHAN, Principal, Sri Venkateswara College of Engineering for

being a source of inspiration throughout the study in this college.

We express our sincere thanks to Dr. N. MEYYAPPAN, Professor and Head,

Department of Chemical Engineering for his permission and encouragementaccorded to carry out this project.

We are also thankful to Dr. C. ANAND BABU, Professor, Department of

Chemical Engineering, Dr. D. SWAMINATHAN, Associate Professor,

Department of Chemical Engineering and Ms. G. HIMA BINDU, Assistant

Professor, Department of Chemical Engineering, Project Coordinators, for their

continual support and assistance.

With profound respect, we express our deep sense of gratitude and sincere

thanks to our supervisor, Ms. G. HIMA BINDU, Assistant Professor,

Department of Chemical Engineering for her valuable guidance and

suggestions throughout this project.

We are very much thankful to Mr. M. MADASAMY, Senior Engineer –

Production and the management of CHENNAI PETROLEUM

CORPORATION LTD for their valuable guidance.

Finally, we thank the faculty and staff of the Chemical Engineering Department,

Sri Venkateswara College of Engineering for their support.



ABSTRACT

CO-PROCESSING OF JATROPHA OIL WITH LINEAR STRAIGHT

RUN DIESEL IN PETROLEUM REFINERY

Jatropha curcas is a species of flowering and fruit bearing plant in the

genus Jatropha. It is cultivated in tropical and subtropical regions around the

world, becoming naturalized in some areas. The seeds contain 27-40% oil

(average: 34.4%) that can be processed to produce a high quality Bio-diesel

fuel, usable in a standard diesel engine. Jatropha oil is an effective alternate

fuel for the engines and equipment that use ‘diesel’ as their fuel.

The cetane number of Jatropha oil is greater than the commercially available

diesel by more than 2 units which helps in low knocking in compressed

ignition engines. Also the sulphur content is low and the inlet temperature of

the reactor or engine for combustion could also be reduced by 1000C with

resultant energy savings.

The technology of Co-processing the Jatropha oil in Diesel Hydro Treating

(DHDT) units of a Petroleum Refinery overcomes the disadvantages of bio-

diesel produced conventionally and produces Bio-diesel with higher Cetane

number, good oxidation stability and lower density. In addition, Co-

processing technology can be deployed in an existing Petroleum Refinery

infrastructure with minor modifications and does not require a separate plant.

This process also costs less as operating cost is reduced by ~ 50% in

comparison to a conventional Biodiesel plant.



TABLE OF CONTENTS

CHAPTER No. TITLE PAGE No.

ABSTRACT ii

LIST OF FIGURES vii

LIST OF TABLES viii

LIST OF SYMBOLS AND ABBREVIATIONS ix

1 INTRODUCTION 1

1.1 History of process 1

1.2 Properties of Jatropha 3

2 AIM & SCOPE 5

3 PROCESS SELECTION AND DESCRIPTION 6

3.1 The new technology ( co-processing) 6

3.2 Process flow diagram 7

3.3 Issues in co-processing technology 8

4 MATERIAL BALANCE

4.1 Material balance around hydrogenator

reactor

10

10

4.2 Material balance around hot separator 14

4.3 Material balance around cold separator 16

4.4 Material balance around scrubber 18

4.5 Material balance around stripper 19

4.6 Material balance around condenser 21

4.7 Overall material balance 22



5 ENERGY BALANCE 23

5.1 Energy balance around feed pre-heater 23

5.2 Energy balance around hydrogenator

reactor 24

5.3 Energy balance around heat exchanger – 1 31

5.4 Energy balance around hot separator 33

5.5 Energy balance around condenser – 1 36

5.6 Energy balance around condenser – 2 37

5.7 Energy balance around stripper 39

5.8 Energy balance around scrubber 40

5.9 Energy balance around cold separator 43

5.10 Energy balance around cooler 44

5.11 Overall energy balance 46

6 PROCESS EQUIPMENT DESIGN 47

6.1.1. Design of reactor 47

6.1.1 Volume of plug flow reactor 50

6.1.2 Volume of packed bed 50

6.1.3 Height of the reactor(H) 51

6.1.4 Diameter of the reactor (d) 51

6.1.5 Height of the packed bed(H) 51

6.1.6 Catalyst used 51

6.1.7 Operating temperature and pressure 51

6.2

DESIGN OF HEATEXCHANGER 52

6.2.1 Film heat transfer coefficient tube side 54

6.2.2 Film heat transfer coefficient shell side 55

6.2.3 % Excess area 57



7 INSTRUMENTATION AND PROCESS CONTROL 59

7.1

INSTRUMENTATION IN WHOLE PROCESS 60

7.1.1 Flow measuring Instruments 60

7.1.2 Temperature Measurement 60

7.1.3 Pressure Measurement 61

7.2

CONTROL SYSTEM FOR HYDROGENATOR REACTOR 61

7.2.1 Reactor feed control 61

7.2.2 Reactor Pressure control 62

7.3

DISPLAY 63

7.4

ALARM 64

7.5

DATA ACQUISITION AND STORAGE 64

8 COST ESTIMATION 65

8.1

Estimation of process equipment cost (pec) 65

8.2

Project cost estimation 66

8.3

Raw material cost 67

8.4

Cost of utilities 68

8.5

Operating cost 69

8.6

Total production cost 70

8.7

Depreciation 70

8.7.1 civil works 71

8.8

Sales 71

8.9

Pay back period 72

8.10

Rate of return 72

9 PLANT LOCATION AND LAYOUT 73

9.1

Factors affecting the layout 73

9.2

Objective of the plant layout 73

9.3

Advantages of scientific layout 74

9.4

Major consideration for layout 74



9.5

Plant layout procedure 78

9.6

Plant layout 79

10 PROCESS SAFETY 80

10.1

Facts about h2s 80

10.2

Hazard 81

10.3

Facts about sulphur dioxide (so2) 82

10.4

Hazard of diesel 82

10.5

Basic rescue procedure 82

10.6

Environment considerations 83

11 SUMMARY AND CONCLUSIONS 84

11.1

Environmental benefits 84

12 REFERENCES 86



LIST OF FIGURES

FIGURE No. TITLE PAGE No.

3.1 Process flow diagram 7

4.1 Material balance around hydrogenator reactor 10

4.2 Material balance around hot separator 14

4.3 Material balance around cold separator 16

4.4 Material balance around scrubber 18

4.5

Material balance around stripper 19

4.6 Material balance around condenser 21

4.7 Overall material balance 22

5.1 Energy balance around feed pre-heater 23

5.2 Energy balance around hydrogenator reactor 24

5.3 Energy balances around heat exchanger-1 32

5.4 Energy balance around hot separator 33

5.5 Energy balance around condenser-1 36

5.6 Energy balance around condenser-2 37

5.7 Energy balance around stripper 39

5.8 Energy balance around scrubber 40

5.9 Energy balance around cold separator 43

5.10 Energy balance around cooler 44

5.11 Overall energy balance 46

6.1 Design of reactor and heat exchanger 47

9.1 Plant layout



LIST OF TABLES

TABLE No. TITLE PAGE No.

4.1 Flow rates of components of stream from hot separator

to cold separator 15

5.1

Values of Standard Heat of Formation 25

5.2

Values of a,b,c,d for H2S 26

5.3

Values of a,b,c,d for H2 26

5.4

Values of a,b,c,d for H2O 29

5.5

Values of a,b,c,d for N2 30

5.6

Values of a,b,c,d for NH3 30

6.1 Design of heat exchanger 50

8.1 Equipment cost 65

8.2 Fixed cost 66

8.3 Raw material cost 67

8.4 Utility cost estimation 68

8.5 Employees salary 69

8.6 Plant and machinery 70



0

LIST OF SYMBOLS AND ABBREVIATIONS

T - Absolute Temperature

m - Mass Flow Rate

Cp - Specific Heat Capacity

ΔT - Difference in Temperature

Q - Heat Energy

ni - Number of Moles of Component ‘i’

ΔHf - Standard Heat of Formation at 298K

DHDS - Diesel Hydrodesulphurisation

DHDT - Diesel Hydrotreating

GHG - Green House Gases

FFA - Free Fatty Acid

FAME - Fatty Acid Methyl Ester

R&D - Research & Development

JO - Jatropha Oil

HSD - High Speed Diesel

MEA - Mono-ethanolamine

PIC - Proportional Integral Controller



1. INTRODUCTION

The need for the technologies to produce alternative, sustainable energy sources,

preferably renewable energy sources has been increasing across the world due to the

concern over limited fossil fuel sources, increasing petroleum prices, government’s

directives and commitment to reduce GREEN HOUSE GASES emissions. Immediate

opportunities to fill the gap in the transportation fuels sector lie in finding the suitable

technologies using renewable resources such as Sugars, Starches, Lignocelluloses, Algal

Oil, Bio-mass, Garbage etc. to produce corresponding compatible fuels. All or many of

these routes have difficulties of cost of production, effect on food chain or demand of a

new infrastructure, which doesn’t attract the present refineries.

Renewable energy sources will fascinate better if the similar fuels to the existing

hydrocarbons are produced to meet the conventional engine technology and compatible to

the existing distribution of infrastructure. Biodiesel [Fatty Acid Methyl Ester, FAME], Di

Methyl Ether [DME] and Synthetic diesel are few alternatives, projected recently in this

direction.

Hydroprocessing of non-edible vegetable oils, rather co-processing of non-edible

vegetable oils with diesel in existing DHDS/DHDT(Diesel hydrodesulphurization/Diesel

hydrotreating) unit will suffice to meet the above discussed issues. Indian Oil [Indian Oil

Corporation Limited] developed a new technology to convert non-edible vegetable oils to

hydrocarbons over indigenous catalyst, which is a simple, sustainable, economical and eco-

friendly process.

1.1 HISTORY OF PROCESS:

Transesterification reaction:

Transesterification process is a well known process to produce bio-diesel. This process

is very complex and capital intensive. Both base and acid catalyzed transesterification

processes are associated with several inherent problems. The Free Fatty Acids (FFA)

present in the vegetable oils interfere with transesterification and deactivate the basic



catalyst. Water deactivates both basic and acidic catalyst, necessitating the drying up of oil.

The soaps formed with basic catalyst from emulsion and soap, which are difficult to

remove. Acids and alkali-based processes produce waste stream and side products resulting

in loss of biodiesel yields.

Enzymatic Transesterification overcomes some of these disadvantages, but the

process is significantly costlier. Super critical transesterification is much simpler among all

transesterification processes. But very high temperature and pressure [3500C and 45MPa]

makes the process less attractive on a commercial scale. Heterogeneous catalysts in this

process have yet to prove their viability. Thus the process though simple suffers from

several disadvantages. The removal of free fatty acids for avoiding saponification in

subsequent steps, removal of product raw glycerin & recovery of excess alcohol used in the

process for complete conversion pose problem during the process.

The bio-diesel produced from the conventional process of transesterification reaction

has several problems, some of which are:

Requirement of removal of free fatty acids for avoiding saponification in

subsequent steps.

Removal and sale of by-product raw glycerin.

Recovery of excess alcohol used in the process for complete conversion.

Water washing is required to remove the caustic employed for neutralization of free

fatty acids & this adds to the plant effluent.

Moreover if the vegetable oil is rancid, an addition step of esterification is

necessary. The process is suitable only for oils having low Free Fatty Acid (FFA)

<0.5%.

Biodiesel has a density of about 0.88 g/cc against the diesel density of 0.825 to

0.845 g/cc and hence, it is difficult to meet the diesel specifications with respect to

density.

It also has a narrow boiling range 3400C+ of which around 20% boils above

specifications of BS-IV diesel i.e. 3600C. Both density and distillation of bio-diesel



requires blending with lighter diesel to meet the diesel specifications, which will

affect the refiner’s profitability adversely.

The presence of oxygen in biodiesel also results in higher emission of NOx. Also,

FAME is not well accepted by the auto industry in all proportions as these are

responsible for injector choking.

These demerits can be overcome by the co-processing technique. Moreover the

properties of Jatropha make it more feasible.

1.2 PROPERTIES OF JATROPHA

It is drought resistant.

It can be grown almost anywhere - even in sandy, saline, or infertile soil.

It adapts well to marginal soils with low nutrient content.

It is relatively easy to propagate.

It is not invasive or damaging.

It is capable of stabilizing sand dunes, acting as a windbreak or combating

desertification.

It naturally repels insects and animals do not browse it.

It lives for over 50 years producing seeds all the time.

It is resilient against the cold.

It does not exhaust the nutrients in the land; rather, it rejuvenates overused

land.

It does not require expensive crop rotation.

It does not require fertilizers.

It grows quickly and establishes itself easily.

It has a high yield.

No displacement of food crops is necessary.

The biodiesel byproduct glycerin is profitable in itself



The waste plant mass after oil extraction can be used as a fertilizer.

The plant itself recycles 100% of the CO2 emissions produced by burning the

biodiesel; two mature plants can absorb 1 metric ton of carbon every year.



2. AIM & SCOPE

The aim of this project is to develop a new alternate for production of bio-diesel by using

the method of co-processing.

100% vegetable oils can be processed by integrating demetallation process in the

DHDT units to get 90-97% paraffinic hydrocarbons and 3 – 10 % aromatics and cyclic

compounds, with a promising life of the catalyst.

On treating the non edible vegetable oil ( Jatropha oil) to remove metals; the process is

well applicable to process even 100 % Jatropha oil, but with little modifications.

Process engineering design also needs little modifications and optimization of the

operating conditions. This is under way at Indian Oil, R&D Center. The product thus

obtained can be blended in regular diesel to meet the existing engine requirements. But

this process needs more metallurgical and catalytic innovations to run sustainable.

Process modifications can be done to produce jet fuels/jet fuel blend stocks.

The process has bright chances to produce lubricants and waxes too.



3. PROCESS SELECTION AND DESCRIPTION

In view of the problems associated with use and production of biodiesel, the refiners

are exploring hydro-processing route as an alternative option. Co-processing results in

improvement in quality of diesel particularly with respect to cetane number and density.

This will enable integrated refining and marketing companies to meet stipulation of

blending bio-fuels in diesel that may be mandated by the government in near future. The

process is capable of handling different vegetable oils & conversions are quantitative.

3.1 THE NEW TECHNOLOGY (CO-PROCESSING)

In this process, triglycerides of fatty acids contained in the vegetable and/or animal oil are

hydrodeoxygenated to form normal C14 to C18 paraffins. Hydrodeoxygenation reactions of

vegetable oils occurs simultaneously with hydrodesulphurization reactions of diesel feed.

Minor cyclization and aromatization to alkyl cyclohexane and alkyl benzene also occur.

The deoxygenation preferably comprises removal of oxygen in the form of water and

carbon oxides from the triglycerides. Hydrodeoxygenation reactions are more exothermic

and also consume more hydrogen in comparison to hydrodesulphurization.

The reactions also remove the Sulphur and Nitrogen content in the oil and hence produce a

more quality diesel.



Oxidation stability of vegetable oils:

Vegetable oil deteriorates during long storage at elevated temperature which can be

avoided by nitrogen blanketing and adding oxidation

stability additives.

CO/CO2 Inhibition:

The CO inhibits the activity of the catalyst by adsorption on active sites, although

inhibition is temporary & less pronounced with NiMo catalyst & higher pressure.

During processing of vegetable oils, water (6.0-8.0% of vegetable processed), CO2

(4.0-6.0% of vegetable processed) and CO (0.5-1.0% of vegetable processed) are

generated depending on the operating conditions and catalyst system. The water can be

separated via boot provided in the separator vessel. The buildup of CO and CO2 in the

recycle gas is therefore required to be controlled. The proper system need to be

installed to reduce CO & CO2 content in recycle gas. Water gas shift & CO methanator

systems may be required. In this process, CO2 was removed by Amine scrubbing and

concentration of CO was controlled by purging of gas from separator.



4. MATERIAL BALANCE

For either total material balance or component material balance we use a simple basic

formula which may be stated as:

Input mass flow rate – Output mass flow rate = Rate of Accumulation

4.1 MATERIAL BALANCE AROUND HYDROGENATOR REACTOR

Hydrogen 330 Kg/hr

Product

28763 Kg/hr

Feed ( JO + HSD)

28433 Kg/hr

Fig. 4.1: Mater ial balance around hydrogenator r eactor

Basis:

30m3/h of Diesel

3m

3

/h of Jatropha oil

Density of feed = 861.6 kg/m3

Density of product = 836.9 kg/m3

Feed Composition in ( wt%)

C = 84.4



H = 13.2

S = 0.6

N = 0.1

O = 1.7

Assuming Overall conversion in the Reactor is 95%.

Hydrogen is supplied in 350% excess.

Basis:

33m3/h × 861.6(kg/m3) = 28432.8(kg/h) = 28433(kg/h)

28433(kg/h) as Feed

Weight of components in (kg/h)

C = 23997

H = 3753

S = 170.598

N = 28.433

O = 483.361

Major chemical reaction:-

H2 + S —˃ H2S (1)

2 + 32 = 34

H2 + 0.5 O2 —˃ H2O (2)

2 + 16 = 18

3H2 + N2 ---˃ 2NH3 (3)

6 + 28 = 34



H2 required for the H2S Reaction (100% conversion)

2/32 × 170.598 = 10.662 Kg/h of H2

H2 required for the H2S Reaction (95% conversion)

10.662 × 0.95 = 10.13 Kg/h of H2

H2 required for the H2O Reaction (100% conversion)

2/16 × 4830361 = 60.42 Kg/h of H2

H2 required for the H2O Reaction (95% conversion)

60.42 × 0.95 = 57.4 Kg/h of H2

H2 required for the NH3 Reaction (100% conversion)

6/28 × 28.433 = 6.09 Kg/h of H2

H2 required for the NH3 Reaction (95% conversion)

6.09 × 0.95 = 5.785 Kg/h of H2

Total H2 required for 95% conversion = 73.314 Kg/h of H2

350% excess H2 supplied = 73.314 × 3.5 = 256.6 Kg/h of H2

Total H2 supplied = 73.314+256.6= 330 Kg/h of H2

H2 Unreacted = 330 - 73.314 = 256.6 Kg/h of H2

Product Stream Calculation:--

For Reaction 1:--

34/32 × 170 598 = 181 26 Kg/h of H2S



For Reaction 2:--

18/16 × 483.361 = 543.781 Kg/h of H2O

For Reaction 3:--

34/28 × 28.433 = 34.526 Kg/h of NH3

For 95% conversion:-

H2S = 181.26 × 0.95 = 172.20 Kg/h

H2O = 543.781 × 0.95 = 516.6 Kg/h

NH3 = 34.526 × 0.95 = 32.80 Kg/h

Unreacted Feed: (5%remains Unconverted)

Nitrogen = 1.42 Kg/h

Sulphur = 8.53 Kg/h

Oxygen = 24.17 Kg/h

Carbon = 23997 Kg/h

Hydrogen = 3753 Kg/h

Hydrocarbons = Hydrogen + Carbon

= (23997+3753) = 27750 Kg/h

At Inlet:

Feed: 28433 Kg/h

H2: 330 Kg/h

Total Inlet: 28763 Kg/h



At Outlet:

H2S: 172.2 Kg/h

H2O: 516.6 Kg/h

NH3: 32.8 Kg/h

Nitrogen: 1.42 Kg/h

Sulphur: 8.53 Kg/h

Oxygen: 24.17 Kg/h

Excess H2: 256.7 Kg/h

Hydrocarbons: 27750 Kg/h

Total Outlet: 28763 Kg/h

Rate of Mass in = Rate of Mass Out

Law of Conservation of Mass is validated.

4.2 MATERIAL BALANCE AROUND HOT SEPARATOR

NH3, H2S, Excess H2, traces of O2 and N2

487.3 Kg/h

Feed from

Reactor Hydrocarbons

28763 Kg/h 277 50Kg/h

Sour Water 525.13 Kg/h



Fig. 4.2: Mater ial balance around hot separator

Basis: 28763 Kg/h of Feed entering the separator

At inlet:

Reactor outlet as feed: 28763 Kg/h


Vapour from Hot separator to cold separator:-

Table 4.2: Flow rates of components of stream from hot separator to cold separator

H2S 172.20 Kg/h

NH3 32.8 Kg/h

Nitrogen 1.42 Kg/h

Oxygen 24.17 Kg/h

Excess Hydrogen 256.70 Kg/h

Total 487.30 Kg/h

At Outlet:

Vapour from hot separator to cold separator: 487.29 Kg/h

Hydrocarbons to Stripper: 27750 Kg/h

H2O as Sour Water

(H2O+Sulphur): 516.6+8.53 = 525.73 Kg/h


Thus, input mass flow rate = output mass flow rate.

Hence, the law of conservation of mass is validated.



4.3 MATERIAL BALANCE AROUND COLD SEPARATOR

H2S, Excess H2, traces of O2 and N2

454.5 Kg/h

Feed

0.05 m3/h

487.3 Kg/h Water spray

H2O + NH4OH 82.8 Kg/h

F ig. 4.3: Material balance around cold separator

Basis: 487.3 Kg/h of Feed entering the Separator

0.5 m3/hr of water is sprayed to remove NH3

Chemical Reaction:

NH3 + H2O ---˃ NH4OH

17 + 18 = 35

Theoretical requirement of water:

18/17 × 32.8 = 34.73 Kg/h of H2O

Water supplied = 0.05 m

3

/hr

Density = 1000 Kg/m3

Thus amount of Water supplied = 0.05 × 1000 = 50 Kg/h

Excess water supplied = 50 - 34.73 = 15.27 Kg/h



At Inlet:

Feed: 487.3 Kg/h

Water supplied: 50 Kg/h

Total Inlet: 537.3 Kg/h

Mass out at Product side as NH4OH

NH3 + H2O ---˃ NH4OH

17 + 18 = 35

Mass flow rate of NH4OH produces (Theoretically)

35/17 × 32.80 = 67.53 Kg/h of NH4OH

Excess water at Product side = 15.27 Kg/h

Total flow rate = 15.27 + 67.53 = 82.8 Kg/h

At outlet:

Vapour from cold separator to Scrubber:

H2S: 172.20 Kg/h

Nitrogen: 1.42 Kg/h

Oxygen: 24.17 Kg/h

Excess Hydrogen: 256.70 Kg/h

Stream to NH4OH: 82.8 Kg/h

Total Outlet: 537.3 Kg/h

Thus, input mass flow rate = output mass flow rate.

Hence, the law of conservation of mass is validated.



4.4 MATERIAL BALANCE AROUND SCRUBBER

Excess H2, <0.01ppm of O2 and N2 to purge & recycle

256.7 Kg/h

Lean amine ( 48.9% wt.)

1226.4 Kg/h

Vapour 454.5 Kg/h

Rich amine with H2S

1424.19 Kg/h

Amine supplied as Lean amine to Scrubber:--

Volumetric flow rate: 1.2 m3/hr

Density of MEA: 1022 Kg/m3

Mass flow rate: 1.2 × 1022 = 1226.4 Kg/h of Lean amine

At Inlet:

H2S: 172.20 Kg/h

Nitrogen: 1.42 Kg/h

Oxygen: 24.17 Kg/h

Excess Hydrogen: 256.70 Kg/h

Vapour as Feed: 454.50 Kg/h

MEA: 1226.4 Kg/h




At outlet:

Hydrogen: 256.7 Kg/h

MEA at outlet stream to other plant: 1226.4 Kg/h

H2S in MEA: 172.2 Kg/h

Oxygen: 24.17 Kg/h

Nitrogen: 1.42 Kg/h


Thus, Rate of Mass in = Rate of Mass Out

Hence, Law of Conservation of Mass is validated.

4.5 MATERIAL BALANCE AROUND STRIPPER:

Lighter Products

4677.2 Kg/h

Hydrocarbons

27750Kg/h

Stripping steam

3567.2 Kg/h

Product Diesel 26640 Kg/h

F ig. 4.5: M aterial balance around str ipper



Basis: 27750 Kg/hr of Hydrocarbons as Feed for Stripper

Specific volume of steam at 1900C =0.157 (m3/Kg) (from steam table)

Density of steam =1/specific volume =1/0.157 = 6.37 (Kg/m3)

Mass flow rate of steam =volumetric flow rate × Density

Mass flow rate of steam =560 × 6.37 =3567.2 (Kg/h)

At Inlet:

Feed (hydrocarbons): 27750 Kg/h

Steam: 3567.2 Kg/h

Total Inlet: 31317.2 Kg/h

Stripper Outlet strips out 4% as Lighter Hydrocarbons:

At outlet:

Diesel Outlet: 27750 × 0.96 = 26640 Kg/h

Lighter Hydrocarbons: 27750 × 0.04 = 1110 Kg/h

Steam Outlet: = 3567.2 Kg/h


Thus, Rate of Mass in = Rate of Mass Out

Hence, Law of Conservation of Mass is validated.



4.6 MATERIAL BALANCE AROUND CONDENSER

Cold water inlet

Off gases

(81.5%)

904.65 + 3567.2

Kg/h

Lighter HC +

Steam

1110 + 3567.2

Kg/h Hot water outlet Naphtha ( 18.5%)

205.35 Kg/h

F ig. 4.6: Material Balance around Condenser

Basis: 1110 Kg/hr of Lighter hydrocarbons entering the Condenser:

At inlet:

Lighter hydrocarbons at inlet: 1110 Kg/h

Superheated Steam at inlet: 3567.2 Kg/h

Total inlet: 4677.2 Kg/h

81.5% of lighter hydrocarbons go as off gases.

At Outlet:

Off gases at Outlet: 0.815 × 1110 = 904.65 Kg/h

Naphtha at Outlet: 0.185 × 1110 = 205.35 Kg/h

Steam: = 3567.2 Kg/h




4.7 OVERALL MATERIAL BALANCE



5. ENERGY BALANCE

5.1 ENERGY BALANCE AROUND FEED PRE-HEATER

HSD (2800C) + JO (250C) + H2 (400C)

9689910.581 kJ/h

Reactor

Outlet, 3700C T H.E.-1 (3050C)

11201347.67 kJ/h 90 90948.832 kJ/h

To The reactor, 3530C

7579511.672 kJ/h

F ig 5.1: Energy Balance around Feed Pre-heater

Hot fluid in = product coming out from reactor

Cold fluid in = Diesel feed + Jatropha oil + hydrogen

Heat energy with cold fluid in = Q with Oil and Q with hydrogen

So,

Qoil = mC p∆T

First let us find out the value of C p of the mixture of Jatropha and diesel feed

C p JO = 0.962 kJ/kg K

C p HSD = 1.3672 kJ/kg K

So average heat capacity = ∑ C pi xi

= 0.962 × 0.1 + 1.3672 × 0.9

= 1.3267 kJ/kg K



Hence,

Qoil = 28433 × 1.3267 × ( 280 - 25 )

= 9619125.581 kJ/h

QH2 = mC p∆T

= 330 × 14.3 × ( 40 - 25 )

= 70785 kJ/h

Total heat energy of cold fluid = 70785 + 9619125.581

= 9689910.581 kJ/h

Heat energy with hot fluid in =Q associated with products from the reactor

Qh = mC p∆T

= 28763 × 1.1288 × ( 370 - 25 )

= 11201347.67 kJ/h

By heat balance we have,

(Qin-Qout)hot = (Qin-Qout)cold

= ( 11201347.67 – 9090948.832 )

= ( 9689910.581 - Qout )

Qout = 7579511.672 kJ/h

5.2 ENERGY BALANCE AROUND HYDROGENATOR REACTOR

Feed, 3530C

7579511.672 kJ/h

Reactor outlet, 3700

C

11201347 97 kJ/h



F ig. 5.2: Energy Balance around Hydrogenator Reactor

We have following three major reactions proceeding inside the reactor

H2 + S —˃ H2S (1)

H2 + 0.5 O2 —˃ H2O (2)

3H2 + N2 ---˃ 2NH3 (3)

We know that Standard heat of reaction =[ ∑ ∆ H f o] products – [ ∑ ∆ H f o] reactants

Table 5.1: Values of Standard Heat of formation

Components H2S O2 N2 H2 NH3 H2O

Standard Heat

of formation

(KJ / mol)

- 20.63 - - - - 46.2 - 285.8

Heat of reaction at any temperature T (∆ HR ) = -∆H1 + ∆ HR o + ∆ H2

Where ∆H1 = heat of formation of reactants

∆H2 = heat of formation of products

∆HR 0 = standard heat of reaction at 298K

For reaction 1:

H2 + S —˃ H2S

To calculate ∆H1:

∆H1 for H2 : = ∫ ni cpi dT

Where ni = moles of H2

Total H2 available for all reactions = 73.314 kmol/h



Total H2 needed for this reaction = 14.663 kmol/h

Here dT = 626 – 298

= 328 K

= 14.663 × 15.2312 × ( 353 - 25 )

= 73253.90 kJ

∆H1 for S : = ∫ ni cpi dT

Since 1 mol of H2 reacts with 1 mol of sulphur,

so moles of sulphur reacting = 14.663 kmol/h

∆H1 = 14.663 × 1.101 × ( 353 - 25)

= 5295.2198 kJ/h

Total ∆H1 for reactants = 73253.90 + 5295.2198

= 78549.1198 kJ/h

To calculate ∆H2 for H2S:

Cp = a + b T + c T2 + d T3

Where a,b,c,d are constants for different components

Table 5.2: Values of a,b,c,d for H2S

A 34.5234

B -17.6381×10-3

C 67.6664 ×10-6

D -53.2454×10-9



∆H1 for H2S = ∫ ni (a + b T2 + CT3 d T3) dT

= [ aT + bT2 / 2 + cT3 / 3 + dT4 / 4]

= 14.663 ×[ 34.5234 (626-298) -

(17.6381×10-3 / 2)

×(6262-2982) + 67.664×10-6 / 3

×(6263-2983) – (53.2452×10-9 /4)

×(6264-2984) ]

= 14.663 [11427.24 - 2672.8071 +

4936.2816 – 1939.2061]

= 172312.3677 KJ/hr

Standard heat of reaction =

∆ HR 0 = -20.63 × 34

= -701.42 kJ

So, heat of reaction for this reaction =

∆ HR1 = -∆H1 + ∆ HR 0 + ∆ H2

= -78549.1198 - 701.42 + 172312.3677

= 93061.8279 kJ/h

For reaction 2:

H2 + 0.5 O2 —˃ H2O

For ∆H1 of H2 = 14.663 × 1.101 × ( 353 - 25)

= 5295.2198 kJ/h

∆H1 for O2 = ∫ ni (a + b T + c T2 + d T3) dT

= ni [aT + bT2 / 2 + cT3 / 3 + dT4 / 4 ]



Number of moles of O2 needed = 0.5 × 14.663 = 7.3315

Table 5.3: Values of a,b,c,d for H2

A 27.0257

B 11.7551 × 10-3

C -2.3426 × 10-6

D -0.5623 ×10-9

= 7.3315 [ 27.0257(626-298)+(11.7551×10-3

/2)

x(6262-2982) – ( 2.3426×10-6 / 3)(6263-2983)

+( – 0.5623×10-9 / 4 ) (6264-2984)

=7.3315(8864.4296 + 1781.3208 - 170.8933 -

20.4790)

=7.3315 [ 10454.378]

= 76646.273 kJ/h

Total ∆H1 = 73253.90 + 76646.273

= 149900.173 kJ/h

∆H2 for H2O = ∫ ni (a + bT +c T2 +d T3) dT

= 1 [aT + bT2 / 2 + cT3 / 3 + dT4 / 4]



Table 5.4: Values of a,b,c,d for H2O

A 32.4921

B .0796 ×10-3

C 13.2107×10-6

D -4.547×10-9

= 14.663 [ 32.4921 (626-298)+ 0.0796×10-3 /2

×(6262-2982)+ 13.2107×10-6 /3

×(6263

-2983

) – 4.547×10-9

/4

×(6264-2984)

= 14.663[10657.4088 + 12.06227 + 963.724 -

165.6025]

= 168149.3099 kJ/h

∆ HR 0 = -285.8 × 18 = -5144.4 kJ/h

So, heat of reaction 2 = ∆ HR2 = -∆H1 + ∆ HR 0 + ∆ H2

= -149900.173 - 5144.4 + 168149.3099

= 148014.7369 kJ/h

For reaction 3:

3H2 + N2 ---˃ 2NH3

∆H1 for H2 = 3 × 73253.90

= 219761.7 kJ/h

∆H1 for N2 = ∫ ni (a + bT +c T2 +d T3) dT

= 14.663 [aT + bT2

/ 2 + cT3

/ 3 + dT4

/ 4]



Table 5.5: Values of a,b,c,d for N2

A 33.1216

B 10.1923 ×10-3

C -4.2621×10-6

D 0.2316×10-9

= 14.663 [33.1216 (626-298) – (10.1923×10-3) / 2

×(6262-2982) - 4.2621×10-6/3

×(6263

-2983

) + (0.2316×10-9

/4)

×(6264-2984) ]

= 14.663[10863.88 - 1544.5003 - 310.9213 + 8.4349]

= 132214.7065 kJ/h

Total ∆H1 for reactants = 219761.7 + 132214.7065

= 351976.4065 kJ/h

∆H2 for NH3 = Cp = a + b T + c T2 + d T3

Where a,b,c,d are constants for different components

Table 5.6: Values of a,b,c,d for NH3

A 25.9796

B -3.9703×10-3

C 24.4989 ×10-6

D 24.2604×10-9



∆H2 for NH3 = ∫ ni (a + b T + c T2 + d T3 ) dT

= 14.663[aT + bT2 / 2 + cT3 / 3 + dT4 / 4]

= 14.633[25.9796(626-298) – (3.9703×10-3) / 2

×(6262-2982) + (24.4989×10-6)/3

×(6263-2983) + (24.2604×10-9 /4)

×(6264-2984) ]

= 14.663[8521.3088 - 601.6434 +

1787.2012+883.5677]

= 14.663 × 10590.4343

= 155287.5381 kJ/h

∆ HR 0 = -46.2 ×34

= -1570.8 kJ

∆ HR3 = -∆H1 + ∆ HR 0

+∆ H2

= -351976.4065 - 1570.8 + 155287.5381

= -198259.6684 kJ

Thus the total heat energy evolved due to endothermicity of the reactions is:

∆ HR1 + ∆ HR2 + ∆ HR3

= 93061.8279+ 148014.7369 -198259.6684

= 42816.8964 kJ/h

This is the total heat gained by the inlet stream.



5.3 ENERGY BALANCES AROUND HEAT EXCHANGER-1

Hot stream from pre-heater, 3050C

9090948.832 kJ/h

Hydrocarbons, 610C 17 00C,

5802649.88kJ/h 1195803 kJ/h

To hot separator, 1930C

4484101.957 kJ/h

F ig. 5.3: Energy Balances around Heat Exchanger-1

We have,

Qh in = mC p∆T

Or, C p = Q/ m∆T

= 9090948.832 / ( 28763 × ( 305 – 25 ))

= 1.1288kJ/kg K

Also,

Qc in = mC p∆T

= 27750 × 1.197 × ( 61 – 25 )

= 1195803 kJ/h

Qc out = mC p∆T

= 27750 × 1.4421 × ( 170 – 25 )

= 5802649.875 kJ/h

Difference in heat = Qc out - Qc in



= 5802649.875 – 1195803

= 4606846.875 kJ/h

Heat gained by the diesel product = Heat lost by the reactor product

Thus,

Heat lost by the reactor product = 4606846.875kJ/h

So the heat along with the hot fluid out = 9090948.832 – 4606846.875

= 4484101.957 kJ/h

Now, the C p of this fluid,

C p = Q/ m∆T

= 4484101.957/ ( 28763 × ( 193 – 25 ))

= 0.9280 kJ/kg K

5.4 ENERGY BALANCE AROUND HOT SEPARATOR

From Condenser 1, 640C , 1439757.665 kJ/h

To Cooler, 590C

To H.E. 1 211426.8 kJ/h

1195803 kJ/h

Sour Water, 400C

32527.865 kJ/h

Fig. 5.4: Energy Balance around Hot Separator



Outlet 1:

Amount of sulphur = 8.53 kg/h

Amount of water = 516.6 kg/h

1.62 % of sulphur + 98.38% of water

C p of sulphur at 400 C= 0.705 kJ/kg K

C p of water at 400 C = 4.186 kJ/kg K

So, C p avg = ∑ xiC pi

= 0.01624 × 0.705 + 4.186 × 0.9838

= 4.1295 kJ/kg K

Heat out = mCp∆T

= 525.13 × 4.1295 × ( 40 – 25 )

= 32527.865 kJ/h

Outlet 2:

Heat out = mCp∆T

= 487.3 × 12.761 × ( 59 – 25 )

= 211426.8 kJ/h

Outlet 3:

Amount of C5H12 = 904.65

So, mass fraction = 904.65 / 27750

= 0.0326

C p of C5H12 at 610C = 1.214 kJ/kg K

Amount of Naphtha = 205.35

So mass fraction = 205 35 / 27750



= 7.4 × 10-3

C p of Naphtha at 610C = 1.983 kJ/kg K

Amount of Diesel = 904.65

So, mass fraction = 26640 / 27750

= 0.96

C p of Diesel at 610C = 1.19 kJ/kg K

C p avg = ∑ xiC pi

= 1.214 × 0.0326 + 1.983 × 7.4 × 10-3 + 1.19 × 0.96

= 1.1967 kJ/kg K

Heat out = mC p∆T

= 27750 × 1.197 × ( 61 – 25 )

= 1195803 kJ/h

Total heat out = Qoutlet 1 + Qoutlet 2 + Qoutlet 3

= 211426.8 + 32527.865 + 1195803

= 1439757.665 kJ/h

Since, Heat in = Heat out

Heat in = 1439757.665 kJ/h

Thus C p of stream coming out from reactor at 640C = Q / m∆T

= 1439757.665 / 28763 × ( 64 – 25 )

= 1.2835 kJ/kg K



5.5 ENERGY BALANCE AROUND CONDENSER-1

From H.E. 1 , 1930C, 4484101.957 kJ/h

Cooling Water, 250C Boiler Feed Water, 860C

3044344.292 kJ/h

To hot Separator, 640C, 1439757.665 kJ/h

F ig. 5.5: Energy Balance around Condenser-1

Qh in = 4484101.957 kJ/h

Qh out = 1439757.665 kJ/h

Difference in Heat energy = Qh in - Qh out

= 4484101.957 – 1439757.665

= 3044344.292 kJ/h

This amount of heat must be carried away by boilr feed water as,

Heat lost by reactor product stream = Heat gained by boiler feed water

Heat gained by boiler feed water = 3044344.292 kJ/h

So, the amount of water needed m = Q /C p∆T

= 3044344.292 / 4.186 × ( 86 – 25 )

= 11922.428 kg/h



5.6 ENERGY BALANCE AROUND CONDENSER-2

LH, 1800C, 2819164.18 kJ/h OffGases,1310C

Water, 250C 1774439.52

kJ/h

Hot Water, 480C Naphtha, 420C

10767.8 kJ/h 8022.203 kJ/h

F ig. 5.6: Energy Balance around Condenser-2

Outlet 1: Lighter hydrocarbons ( C5H12) + Water vapour

QLighter hydrocarbons = mC p∆T

= 904.65 × 1.667 × ( 131 – 25 )

= 159853.46 kJ/h

Qwater = mC p∆T

= 3567.2 × 4.27 × ( 131 - 25 )

= 1614586.06 kJ/h

Outlet 2: Naphtha

Qnaphtha = mC p∆T

= 205.35 × 2.298 × ( 42 – 25 )

= 8022.203 kJ/h



Total Qout = 159853.46 + 1614586.06 + 8022.203

= 1782461.7 kJ/h

Inlet : Superheated Steam + Lighter Hydrocarbons ( C5H12 + Naphtha)

Qsteam = mC p∆T

= 3567.2 × 4.42 × ( 180 – 25 )

= 2443888.72 kJ/h

QLC = mC p∆T

= 1110 × 2.1812 × ( 180 – 25 )

= 375275.46 kJ/h

Difference in heat = Qin - Qout

= ( 2443888.72 + 375275.46 ) –

( 159853.46 + 1614586.06 + 8022.203)

= 2819164.18 – 1782461.72

= 1036702.46 kJ/h

This amount of heat must be carried away by cooling water as,

Heat lost by reactor product stream = Heat gained by cooling water

Heat gained by cooling water = 1036702.46 kJ/h

So, the amount of water needed m = Q /C p∆T

= 1036702.46 / 4.186 × ( 48 – 25 )

= 10767 8kg/h



5.7 ENERGY BALANCE AROUND STRIPPER

LH, 1800C,

Hydrocarbons, 1700C 2819164.18KJ/hr

6487761.7 kJ/h

Steam, 1900C Product Diesel, 1600C

2625102.48 kJ/h 6293700.00 kJ/h

F ig. 5.7: Energy Balance around Str ipper

Heat outlet:

Outlet 1: Diesel product

Q = mC p∆T

= 26640 × 1.75 × ( 160 - 25 )

= 6293700.00 kJ/h

Outlet 2: Water vapour + Naphtha + C5H12 = Inlet for condenser 2

So, Q = 2819164.18 kJ/h

Total Qout = 6293700.00 + 2819164.18

= 9112864.18 kJ/h

Inlet 1: Superheated steam 1900C

Qsteam = mC p∆T

= 3567.2 × 4.46 × ( 190 – 25 )



= 2625102.48 kJ/h

Inlet 2: Hydrocarbons

We have, Qout = Qin

So, 9112864.18 = QHC + 2625102.48

Thus, QHC = 9112864.18 – 2625102.48

= 6487761.7 kJ/h

The heat energy associated with this stream is same as that coming out from heat

exchanger 1.

5.8 ENERGY BALANCE AROUND SCRUBBER

Hydrogen, 400C 55062.15 kJ/h

Lean amine, 380C

From cold separator, 390C 44306.1528 kJ/h

68632.7972 kJ/h

Sour amine, 410C

57876.8 kJ/h

F ig. 5.8: Energy Balance around Scrubber

Inlet 1: 48.9% Monoethanolamine

Q = mC p∆T

= 1226.4 × 2.779 × ( 38 – 25 )



= 44306.1528 kJ/h

Outlet 1: Hydrogen

Q = mC p∆T

= 256.7 × 14.3 × ( 40 – 25 )

= 55062.15 kJ/h

Outlet 2: MEA + H2S + O2 + N2

Amount of Monoethanolamine = 1226.4 kg/h

Mass fraction of MEA = 1226.4 / 1424.19

= 0.861

C p of MEA at 410C = 2.779 kJ/ kg K

Amount of H2S = 172.2 kg/h

Mass fraction of H2S = 172.2 / 1424.19

= 0.1209

C p of H2S at 410C = 1.003 kJ/ kg K

Amount of O2 = 24.17 kg/h

Mass fraction of O2 = 24.17/ 1424.19

= 0.01697

C p of O2 at 410C = 0.920 kJ/ kg K

Amount of N2 = 1.42kg/h

Mass fraction of N2 = 1 42/ 1424 19



= 9.9705 × 10-4

C p of N2 at 410C = 10.4 kJ/ kg K

C p avg = ∑ xi C pi

= 0.861 × 2.779 + 0.1209 × 1.003 + 0.01697 × 0.920

+ 9.9705 × 10-4 × 10.4

= 2.5399 kJ/kg K

So, Q = mC p∆T

= 1424.19 × 2.5399 × ( 41 – 25 )

= 57876.8 kJ/h

We have, Qout = Qin

57876.8 + 55062.15= Qcold sep + 44306.1528

Qcold sep= 112938.95 – 44306.1528

= 68632.7972 kJ/h

C p of stream coming out from cold separator = Qcold sep / m∆T

= 68632.7972 / (454.5 × ( 39 – 25 ))

= 10.786 kJ/kg K



5.9 ENERGY BALANCE AROUND COLD SEPARATOR

To scrubber, 390C , 68632.7972 kJ/h

From cooler, 380C Spray water, 280C

71809.337 kJ/h 627.9 kJ/h

Water + NH4OH , 360C , 3804.44 kJ/h

F ig. 5.9: Energy Balance around Cold Separator

Outlet 1: Stream to scrubber

Q = 68632.7972 kJ/h

Outlet 2: Water + Ammonium hydroxide

QH2O = mC p∆T

= 15.27 × 4.186 × ( 36 – 25 )

= 703.12 kJ/hr

Q NH4OH = mC p∆T

= 67.53 × 4.175 × ( 36 – 25 )

= 3101.32 kJ/hr

Total = 703.12 + 3101.32

= 3804.44 kJ/hr



Inlet 1: Spray water

QH2O = mC p∆T

= 50 × 4.186 × ( 28 – 25 )

= 627.9 kJ/hr

We have,

Qout = Qin

3804.44 + 68632.7972 = Qcooler + 627.9

Qcooler = 71809.337 kJ/hr

C p of Stream coming out after cooler = Qcooler / m∆T

= 71809.337 / ( 487.30 × ( 38 – 25 )

= 11.336 kJ/kg K

5.10 ENERGY BALANCE AROUND COOLER

To cold separator , 380C, 71809.337 kJ/h

Water , 250C

Hot water, 490C ,

139617.46KJ/hr

From hot separator , 590C, 211426.8 kJ/hr



F ig. 5.10: Energy Balance around Cooler

Qout = 71809.337 kJ/hr

Qin = mC p∆T

= 487.3 × 12.761 × ( 59 – 25 )

= 211426.8 kJ/hr

Difference in heat = 211426.8 – 71809.337

= 139617.463 kJ/hr

This heat must be carried away by the cooling water used for cooling

Amount of water required m = Qdiff / C p∆T

= 139617.463 / 4.186 × ( 49 – 25 )

= 1389.72 kg/hr



6. PROCESS EQUIPMENT DESIGN

DESIGN OF HYDROGENATOR REACTOR:

Collect the availability data

Write the general reaction and find no of moles at t=0 t=t

Find Cso and Fso , Fso =sulphur inlet flow rate/mol.wt

Find out the conversion factor X= Fso - Fsf / Fso

Design the equation for the plug flow reactor

Derive the relationship for volume of the reactor and space time

Calculate the diameter and height of the reactor using thumb rule

Available data:

1.feed (diesel+oil) flow rate rate:33 m3/hr

2.mafke up H2 floiw rate : 330 kg/hr

3.Initial sulfur content present in diesel: 33000ppm(3.3%)

4. final sulfur content present in diesel: 12500ppm(1.25%)

PHYSICAL PROPERTIES OF DIESEL AND HYDROGEN

Inlet temperature : 353 ℃

Inlet pressure: 69.4kg/gcm2

Space time : 0.9127hr

Catalyst used is Ni-MO

REACTION: HC-S+H2 →H2S+HC

S.N0 Components No of moles at t=0 No of moles at t=t

1. HC-S a a(1-x)

2. H2 b b(1-x)

3. H2S 0 Ax

4. HC 0 Bx



-r s =K (HC-S)n

-r s =K (a(1-x))n

= K (an(1-x))n

Cso = 0.033 kg of sulfur/kg of diesel

= 0.033/32 kg of sulfur/ kg of diesel

=10.312510^-4 kmol of sulfur/ kg of diesel

= 0.888525 kmol of sulfur/ m3 of diesel

sub the value of Cso in equation in (1)

-rs= k (0.888525)^ 0.5 (1-x)^ 0.5

-rs= k*0.943*(1-x)^0.5

Diesel flow rate = 33* 861.6

= 28432.8 kg/hr

Sulfur inlet flow rate = 28432.8* 3.3/100

= 938.28 kg/hr

Fso= 938.28/32

= 29.321 kmol/ hr

106kg of diesel contain 300 kg of sulphur

28432.8 kg/hr diesel contain y kg of sulfur

Y=300 × 28432.8/10^6

= 8.5298 kg/hr



Fsf = 92.598/ 32 = 0.2666 kmol/hr

Conversion X = Fso- Fsf / Fsf

= Fso-Fsf/ Fso

= 29.321- 0.2666/ 29.321

= 0.99(99%)

We know that for the plug flow reactor the design equation

=

= ∫

−

V=Fso∫

−

Sub value of – in equation

V=

×.43 ∫

(−)ˆ/2

For finding the value of ∫

(−)ˆ/2

Take X=cosθ

=-sinθ

∫− in d

(−)ˆ/2

1-cos θ=sin2

2 +cos2

2 -[cos2

2 – sin2

2]

=2sin2

2

Sinθ=2sin

2 cos

2

∫− in d

(−)ˆ/2

=∫

−2in

2

in2

=-21/2∫ cos

2θ

= -2^1/2 *2(sin

2) 8.1096

90

= -2* 2^ ½ ( 0.071-0.707)



∫

(−)ˆ/2

= 1.799

Sub equation (4) in equation (3)

V=

×.43

× 1.799

V= 29.321* 1.799/ k* 0.943

= 55.937/ k

We know that

τ= Cso ∫

−

τ= 0.9127 hr ^-1

0.9127 = 0.888525/ k* 0.943 ∫

(−)ˆ/2

* 1.799

K = 1.8572

Substitute k value in equation (5)

V = 55.937/ 1.8572

= 30.119 m3

Π* d2h/4 = 30.119 m3

According to Thumb rule

h/ d =4

so, h=4d

substitute in equation (6)

π* d2 (4d) /4 = 30.119

d3 = 9.5872

d = 2.1244 m

h = 4d

= 4 * 2 1244



= 8.497

H = 8.50 m

Volume of bed = 0.8 ( volume of the reactor)

= 0.8 ( 30.119)

= 24.0952 m3

π * d2 * H/ 4 = 24.0952

π* (2.124)^2* H/4 = 24.0952

H = 6.8 m

Height of the regenerator reactor = 6.8 m

DESIGN SUMMARY:

1. Volume of the plug flow reactor = 30.119m3

2. Volume of the packed bed = 24.0952m3

3. Height of the reactor (h) = 8.5 m

4. Diameter of the reactor (d) = 2.1244m

5. Height of the packed bed (H) = 6.8m

6. Catlyst used = Ni-Mo

7. Operating temperature = 353 ℃

8. Operating pressure = 69.4 kg/ cm2



DESIGN OF HEAT EXCHANGER

ITEM DESCRIPTION UNIT SHELL SIDE TUBE SIDE

Fluid circulated - Hydrocarbon Reactor efflue

Fluid flow rate kg/s 7.74 7.799

Fluid temperature Inlet/outlet ºC 61/170 305/193

Specific heat kJ/kg ºC 1.197 1.1288

Viscosity Cp 0.451 0.39

Density kg/m3 959 861

Pressure drop allowable kPa 70 70

Thermal conductivity

Combined dirt factor

W/m ºC 0.66 0.081

m2 ºC/W 7.9 * 10-4

Tube dimensions = 20mm OD × 15.8mm ID ×4880 mm long

Tube pitch = 25 mm triangular pitch

Heat Duty

Heat load on shell side fluid, Q = mh * Cph * ΔTh

where,

Q = Heat transferred, kW

mh = Mass flow rate of hot fluid, kg/s

Cph = Specific heat capacity of hot fluid, kJ/kg ºC

ΔTh = Temperature difference, ºC

Q = 7.799 * 1.1288* (305-193) =1010.1 kW

Q = 1010.1 kW

Flow rate of cold fluid

Heat load on tube side fluid, Q = mc * Cpc * ΔTc

Q = mc * Cpc * ΔTc = mh * Cph * ΔTh

1010.1 = mc (1.197)(170-61)

mc = 7.74kg/s



Driving force for heat transfer

305 hot fluid- reactor effluent 193

170 cold fluid – hydrocarbon ← 61

LMTD =∆− ∆2

∆∆

=3− 32

=3

.22 = 133 ℃

LMTD Correction factor

Temperature Efficiency, P =∆ℎ

∆ =

(7−6)

(3−6) = 0.447

Heat capacity rate ratio , R =∆ℎ

∆ =

(3−3)

(7−6) = 1.03

From correction factor chart FT is found to be 0.95

Corrected mean temperature difference ΔTm = FT × LMTD

= 0.95× 133= 126.4ºC

Area Required for Heat Transfer

Assuming the heat transfer coefficient value as Uo = 500 W/ m2 ºC

We have

Q = Uo Ao ΔTm

Where,

Q = Total heat to be transferred per unit time, kW

Uo = The overall heat transfer coefficient, W/m2 ºC

Ao = Heat transfer area required, m2

ΔTm = Corrected mean temperature difference, ºC

Heat transfer area required Ar =

×∆ =

. ×^3

×26.4 = 15.98 m2

To find number of tubes

Tube dimensions = 20mm OD × 15.8mm ID ×4880 mm long

Tube pitch = 25 mm triangular pitch

Surface area per tube = Π DoLe

= Π×20×10-3×(4880-2×25) ×10-3



= 0.303m2

No of tubes NT =Ttal heat tranfer area reqd

Surfae area per tube

=.8

.33

= 52.7 ≅ 53 tubes

From the tube count table the heat exchanger unit selected will have 254 mm (10”) shell

ID with 53 tubes, 1 – shell & 4 – tube passes.

Determination of Film Heat Transfer Coefficient

Film Heat Transfer Coefficient Tube Side Fluid (hio)

hiDi = 0.023 (Re)0.8(Pr)0.33(µ

µ)0.14

k

where

hi = Film heat transfer coefficient inside the tubes, W/m2 ºC

Di = Inside diameter of the tubes, m

K = Thermal conductivity, W/m ºC

Re = Reynolds’s number = GtDi / µ

Pr = Prandtl number = Cp µ / k

Cp = Specific heat capacity, kJ/kg ºC

Gt = Mass velocity kg/s. m2 = mc/ At

mc = Mass flow rate of cold fluid, kg /s

At = Tube side flow area per pass, m2

μ = Viscosity of cold fluid, kg/m.s

μw = Viscosity with respect to tube wall temperature, kg/ms

At = ( Π/4) × Di2 (NT/NP)

= ( Π/4) × (15.8×10-3)2×(3

4 ) = 0.0030 m2

Gt = ( mc / At) =7.74

.3= 2580 kg/m2s

Tube side Re = GtDi / µ =28×.8

.3×. = 104523



Pr = (Cp µ / k) =.288××.3×.

.8 = 5.435

hiDi = 0.023 (104523)0.8(5.435)0.33(1)0.14

k

hi = (416.59×0.081) / (0.0158)

hi = 2135.68 W/m2 ºC

hio = hi *(

) = 2135.68 * (0.0158/0.02) = 1687.19 W/m2 ºC

hio = 1687.19 W/m2 ºC

Film Heat Transfer Coefficient Shell Side (ho)

hoDe = 0.36 (Re)0.55(Pr)0.33(µ

µ)0.14

k

where

ho = Film heat transfer coefficient inside the tubes, W/m2 ºC

De = Effective diameter, m

k = Thermal conductivity, W/m ºC

Re = Reynolds’s number = GsDe / µGs = Mass velocity kg/s. m2 = ( ms / As)

ms = Mass flow rate of hot fluid, kg /s

As = Shell side cross flow area, m2 = ( DsC’B) / PT

μ = Viscosity of hot fluid, kg/m s

Pr = Prandtl number = (Cp µ / k)

Cp = specific heat capacity, kJ/kg ºC

μw = viscosity with respect to tube wall temperature, kg/m s

For triangular pitch

De = {4 ( 0.86PT2

4 DO

2) } / DO

De =4{(.86×.2×.2)− (

)×.2×.2}

×.2 = 0.0142 m

Mass velocity GS = (mS / AS)

Shell side Cross Flow Area , AS = ( DSC’B) / PT



where

Ds = shell ID

C’ = clearance = PT – OD

B = baffle spacing = (1/5) DS to DS

B = 0.5 Ds , Ds = 0.254m

As = ( DsC’B) / PT = ( 0.254×0.005×(0.5×0.254)) / 0.025 = 0.00645 m2

Gs = (ms / As) = (7.74/ 0.00645) = 1200 kg/m2s

Shell side Re = GsDe / µ =2×.42

.4×. = 37782.71

Pr = (Cp µ / k) =.7××.4×.

.66 = 0.818

ho(0.0142) = 0.36 (37782.71)0.55(0.818)0.33(1)0.14

0.66

ho = 5155.6W/m2 ºC

Determination of Wall Resistance

Wall Resistance , R W = DO ln (DO/Di)

2K W

= 0.02 ln (0.02/0.0158) ( K W = 50 W/m2 ºC)

2×50

= 4.71×10-5 m2 ºC/ W

Combined dirt factor

R do + R dio = 5.9 ×10-4 m2 ºC/ W

Overall heat transfer Coefficient, UO

UO = [ (1/ ho) + R do + R dio + R W + (1/ hio) ] -1

= [ (1/ 1687.19) +7 .9×10-4 + 4.71×10-5 + (1/ 5155.6) ] -1

UO = 515.85W/m2 ºC



Heat transfer area Required , Ar

Ar =

×∆ =

. ×^3

.8×26.4 = 15.5 m2

Heat transfer area available in the assumed Unit, Aa

Aa = Π DoLeNT

= Π×20×10-3×(4880-2×25) ×10-3× 53

= 16.08m2

% EXCESS AREA = { (Aa Ar )*100}/ Ar

= {(16.08 – 15.5)*100}/15.5 = 3.74%

PRESSURE DROP CALCULATIONS

Pressure Drop Tube Side (ΔPT)

ΔPT =

2×× ×× + (2.5× N P ×V 2×S), kPa

where,

f = Friction factor = 0.72 (Re)-0.33

Gt = Mass velocity, kg/s.m2

L = Length of the tubes

NP = Number of tube passes

Di = Inside diameter of tubes

= (µ

µ)0.14 = 1

S = specific gravity = 0.861

f = 0.72 * (104523)-0.33 = 0.016

v = Gt / ρ = (258/861) = 0.30m/s

ΔPT = .6×28

×4.8×4

2××.8×.86× + (2.5×4× 0.3 2×0.861), kPa

= 7.5+0.77 = 8.27 kPa

Tube Side Pressure Drop = 8.27 kPa



Pressure Drop Shell Side (ΔPS)

ΔPS =

(+)

2×× ××, kPa

where,

f = Friction factor = 1.87 * (Re)-0.2 = 1.87*(37782.7) -0.2 = 0.227

GS = Mass velocity, kg/s.m2 = 1200 kg/m2s

S = Specific gravity = 0.957

DS = Shell ID = 0.254m

N b = No. of baffles = ℎ

=4.83

.×.24

= 38 baffles

De = Equivalent diameter, m = 0.0142m

ΔPS =.227×2

×.24(38+)

2××.42××.7, kPa

= 19.62 kPa

DESIGN SUMMARY

Heat duty of the heat exchanger = 1010.1 kW

Area required for heat transfer = 15.5 m2

Area available for heat transfer = 16.08 m2

% Excess area =3.74 %

Number of tubes = 53

Number of passes on tube side = 4

Number of passes on shell side = 1

Shell ID = 254mm

Film heat transfer coefficient tube side , hio = 1687.19 W/m2 ºC

Film heat transfer coefficient shell side , ho = 5155.6 W/m2 ºC

Pressure drop tube side ΔPT = 8.27 kPa

Pressure drop tube side ΔPS =19.62 kPa



7. INSTRUMENTATION AND PROCESS CONTROL

Instrumentation and process control engineering plays vital role in controlling the

process for desired output from the plant. All instruments should be installed from standard

manufacturers by selecting based on the previous process experience.

Instrumentation is provided to monitor the vital process parameters during plant

operations. They may be incorporated in automatic control loops or used for the manual

monitoring of process parameters. They may also be a part of automatic computer data

acquisition. Instruments monitoring critical process parameters will be fitted with

automatic pop-ups and enunciations to alert the operator at critical and hazardous situations

in prior. The main objective of the designer when specifying instrumentation and control

schemes are, to keep process parameters within the operating limit, so as to detect

dangerous situations that may develop due to process deviations and to provide

alarms/buzzer and automatic fail safe systems. And also to maintain the product

composition within the specified quality standards and to operate the plant at the lowest

production cost. Variables in a process must be measured and then controlled and

integrated for optimum processing conditions. Mechanical and electrical components and

systems have been designed by instrumentation and process control engineers to reduce

labor and improve feasibility of plant operation, and to allow extensive outdoor plant

construction.

The economic advantage of automatic process control has been well established

throughout the industry, thus accounting for the rapid growth in this area of the chemical

process industry.

The main objectives of designer when specifying instrumentation and process

control are:

Monitor and control the process parameters within the operable range on real time

basis.

Maintaining the product quality within the specifications.

Ensure optimum production cost with integrated fail safe mechanisms.



Eliminating all the pit-falls that may rise due to manual intervention and properly

taken care of such situations and alert by prior signaling and controlling

catastrophic failures automatically before-hand.

7.1 INSTRUMENTATION IN WHOLE PROCESS

7.1.1 Flow measuring Instruments

For measuring the flow of diesel feed from the distillation column Mass flow meter

will be used, it is precious and sensitive measuring instrument it will give accurate and fast

feed back to the control system. Mass flow meters are available in various ranges; in this

process stainless casing is suitable to get good results. After heating the Tetrahydrofuran to

vapor form the mass flow meter is not suitable as it has a very high thermal expansion co-

efficient, at the measurement of vapor flow of THF differential pressure orifice meters are

suitable for flow measurement.

For measuring the flow of Jatropha oil feed to the reactor the same mass flow

meters can be used as the co-processing percentage is must to be kept specific.

For measuring the flow of H2 feed to the reactor differential pressure orifice meters

are generally using. In this instrument concentric orifice plate is installed in the pipeline to

drop the pressure of the feed, this pressure drop is directly proportional to the square of the

velocity. Flow rate can be obtained manipulating the velocity with the area of the orifice,

sometimes the vapor pressure taping lines of the orifice meter requires flushing to avoid

the formation any condensate in the line.

7.1.2 Temperature measurement

For measuring the temperature of reactants, resistance thermometers or

Thermocouples are used. Platinum- platinum rhodium thermocouples are used in most

industries, since thermocouples are gives excellent transmission accuracies than other

temperature measuring instruments. THF temperature is measured by thermocouples

mentioned.

For measuring the temperature of Reactor the same thermocouples are used.



7.1.3 Pressure Measurement

The bourdon tube measuring elements are used in combination with the differential

pressure transmitters with stainless steel internal components are used for the measuring

the reactor pressure drops in mm water column.

7.2 CONTROL SYSTEM FOR HYDROGENATOR REACTOR

For the reactor the following controls are required for safe and efficient operation

of the reactor.

Raw materials flow controlling as per the specified molar ratio.

Reactor inside pressure controlling by varying the outlet gas flow.

In the hydrogenator, raw materials react to produce the diesel associated with water.

The conversion in the reactor is very much dependent on the molar proportion of

the reactants. The online molar feeds are measured and the controlled for the

specified mole ratio of diesel feed : Jatropha oil feed to get the desired output from

the reactor. The required pressure drop through the reactor for consistent plug

conditions is supplied by the compressor discharge pressure and is maintained by

the reactor outlet damper.

7.2.1 Reactor feed control

The feed rates of the reactants to the reactor are usually controlled to maintain the

proper throughput of the reactor. If there is only one feed flow, it may be controlled with

simple flow control. A special type of flow control, Ratio Controller, will be used. The

molar ratios of Diesel feed and Jatropha oil will be controlled by Ratio Controller, with this

controller the ratio of the individual feeds must often be maintained in order to control the

amount of excess reactants left in the out let stream of the reactor, to control the

temperature, or to favor the production of Diesel. Ratio Controller can be used to maintain

the proper feed ratio. The ratio may be set by the operator, either as a fixed value or as a

value modified by off line analysis of the reactor product stream. It may also be modified

online by a temperature controller or composition controller.



In the flow control loop the dynamic elements consist of the controller, the flow

measuring element and the control valve. The control action for a flow control system is

generally PI. The integral control is needed to eliminate offset and thereby establish a

precise ratio of the mixed streams. Derivative action is usually avoided in flow control

because the signal from a flow measuring element is inherently noisy. The presence of

derivative action would amplify the noise and give poor control.

7.2.2 Reactor Pressure control

Pressure is another key process variable because pressure provides a critical

condition for the reactor. Poor pressure control can lead to major safety, quality and

productivity problems. High pressure inside the reactor can cause explosion. Therefore it is

highly desired to keep pressure inside the reactor in good control. The pressure in the

reactor is controlled by PIC controller which includes a low and high alarm. Addition to

this a high pressure switch is provided with bypass key for the safety of the reactor. Bypass

key are useful to carry out any maintenance jobs on the switches. Field mounted pressure

gauge is provided on top of the reactor.

Interlocks are control functions that prevent the normal operation of other control

or operating functions. They are used to ensure personnel or equipment safety or to protect

the operability of the process. For example, in order to protect the pumps, interlocks will

often be used to prevent the pump from operating if a valve is closed on either suction or

the discharge side of the pump or if the level of the tank feeding the pump is low.

Interlocks that do not take intermediate action, but prevent either the operator or a control

function from operating, are often referred to as permissive. If interlocks are required for

personnel safety, they should normally be separate from the regulatory control system in

order to protect against control system failure.

The reactor interlock functions are cut off feed by closing valves and stopping the

H2, if the reactor temperature or pressure is high. Enforce startup sequences by preventing

the introduction of feeds until proper conditions are met.

Because of the instability of the exothermic reactions and the possibility of a

runaway reaction, the reactor is often equipped with an emergency shutdown system, as

well as other safety equipment, such as burst diaphragms and release containment systems.

The purpose of the emergency shutdown system is to stop the reaction in the event of a

runaway The system may accomplish this by quickly reducing the temperature injecting



some material into the reactor, which will reduce the rate of the reaction; or venting the

reactor to reduce the pressure. The emergency shutdown system must be highly reliable.

In almost all industries Computer based Distribution control systems and

Programmable logic controllers with human interface systems are installed for easy control

through computers, in this system the hardware requirements for pneumatic, electronic,

and microprocessor-based controllers are working, all pneumatic signals in the range of 3

to 15 psig , the energy needed to operate these pneumatic components is a source of clean,

dry air at a pressure of about 20 psig. The pressure can vary from 20 psig by about +-10 %

without adversely affecting the operation of the instruments.

The electronic system required both electrical and pneumatic power to operate the

components. A transducer or converter is needed between the controller and the valve to

convert current (4-20 mA) to pressure (3- 15 psig)

The components for a microprocessor based system contains the control algorithm

reside as a computer program in the memory of the computer. The operator communicates

with the control system with a keyboard, a monitor, and a printer. The computer can

perform many more functions than implementation of the control algorithm; the recorder

of the pneumatic or electronic system is replaced by a monitor screen on which transients

are shown.

In a modern controller both analog and digital signals are processed. The analog

signal is the type that represents a continuous variable with varies over a range of values.

The digital signal is a binary signal that can be represented by two states.

7.3 DISPLAY

The software in a modern controller has made the strip chart recorder almost

unnecessary. Through the use of skilled programming, the transients produced in a control

system can be displayed on a monitor screen dynamically. As time progresses, the values

of selected variables are displayed as a function of time. The segment time shown on the

screen can be selected to be a few minutes to a few hours to show dynamic detail or long

term trends. Transients that occurred in the past can be stored and displayed again.



7.4 ALARM

An extensive amount of the software in modern controllers is devoted to detecting

and reporting a problem in the form of an alarm. The alarm takes the form of a visual

signal, an audible signal (beeping horn), or the auction of a switch.

7.5 DATA ACQUISITION AND STORAGE

Long term storage of the transients can be obtained easily with a digital computer;

this task is referred to as archiving. The automatic storage of critical process control

variables on disk or tape can be retrieved later to explain process operating difficulties. The

computer can also be used to automatically record or log the type and location of an alarm,

the time of a process alarm, the time of acknowledgement of an alarm, and the time it wascleared by operator intervention.



65

8. COST ESTIMATION

8.1 ESTIMATION OF PROCESS EQUIPMENT COST (PEC)

Table 8.1: Equipment Cost

S. No Equipment Quantity Cost( Rs in Lakh)

1 Jatropha oil tank 1 3.4

2 Diesel feed tank 1 6

3 Hydrogenator 1 112.5

4 Separators 2 42

5 Stripper 1 34

6 Scrubber 1 43.6

7 Condenser 1 37.4

Total estimated cost of Equipment= Rs 278.9 Lakhs



66

8.2 PROJECT COST ESTIMATION:

Table 8.2 Fixed Cost

S. No Descriptions Cost (Rs in Lakhs)

1 Plant and Machinery cost 278.9

2 Installation cost (35% PEC) 97.62

3 Instrumentation cost (35% PEC) 97.62

4 Piping and pumping cost (60% PEC) 167.34

5 Electrical cost (45% PEC) 125.51

6. Land cost (Total Area 8 acres @ Rs 200

Lakhs / acre)

1600.00

7. Service facilities (50% PEC) 139.45

Total Fixed Cost = Rs 2506.44 Lakhs

Indirect Cost

Construction expenditure (50% of purchase cost) = 139.45 Lakhs

Engineering & Supervision (30% of purchase cost) = 83.69 Lakhs

Total Indirect Cost = 223.17 Lakhs

Total Direct and Indirect Cost = 2729.61 Lakhs

Contingency (10% of Direct and Indirect Cost) = 272.96 Lakhs

Contractors fees (10% of Direct & Indirect cost) = 272.96 Lakhs

Total Fixed Capital Investment = 3275.53 Lakhs

Working Capital (25% of Fixed Capital Investment) = 818.88 Lakhs



67

Total Capital Investment = Fixed Capital Investment

+ Working Capital

= 3275.53 + 818.88

= 4094.41 Lakhs

8.3 RAW MATERIAL COST

Basis: On stream Days 330 Considered

Table 8.3 Raw Material Cost

S.No Raw Material Cost ( Rs) Quantity m3

(Per year)Cost (Rs in Lakhs)

1 Boiler feed water 136 per m3 28648.224 38.96

2 Jatropha oil 80 per kg 22518936 Kg 18015.15

3 Diesel feed 47 per kg 202670424 Kg 95255.10

Raw Material Cost = Rs 113309.21 Lakhs



68

8.4 COST OF UTILITIES:

Table 8.4: Utility Cost Estimation

S. No Utility Cost ( Rs) Quantity (Per year) Cost (Rs in Lakhs)

1 Electricity

(555.55KWh)

Rs 10 /KWh

(Power taken

from Refinery)

555.55KWh×24hr×

330days

=4399956KWh

439.98

2 L.P Steam 1867 per Ton 35640 Ton (-) 655.40

3 M.P Steam 1926 per Ton 43560 Ton (-) 838.96

4 Water 65 per m3 565488 367.56

Total = Rs (-) 686.82 Lakhs

Cost Of Utility = Rs (-) 686.82 Lakhs

(-) Indicates NET EXPORT to Other Unit



70

8.6 TOTAL PRODUCTION COST

Basis: On-stream Days 330 Considered

Raw Material Cost = Rs 113309.21 Lakhs

Utilities Cost = Rs (-) 686.82 Lakhs

Maintenance Cost = Rs 100 Lakhs

Salary and Wages = Rs 185.20 Lakhs

Catalyst and Chemicals = Rs 55.00 Lakhs

Total Production Cost = Rs 112962.59 Lakhs

8.7 DEPRECIATION

Depreciation Rate = 5.25% per annum

Table 8.6: Plant and Machinery

S. No. Equipment Cost (Rs in Lakhs)

1 Plant and Machinery cost 278.9

2 Installation cost (20% PEC) 55.78

3 Instrumentation cost (15% PEC) 41.84

4 Piping and pumping cost (20% PEC) 55.78

5 Electrical cost (20% PEC) 55.78

Total = Rs 488.08 Lakhs

Depreciation for Plant and Machinery = 0.0525 × 488.08

= Rs 25.62 Lakhs



71

8.7.1 Civil Works

Depreciation Rate = 2.50% per annum

Depreciation Amount = 0.025 × 227.9

= Rs 5.69 Lakhs

Total Depreciation = Rs 31.31 Lakhs

8.8. SALES

Diesel production in One day = 639.36 MT

Onstream Days Considered = 330 Days

Production in One year = 639.36 × 330

= 210988.8 MT

Selling Price = Rs 55000 per MT.

Sales Revenue in One year = Rs 116043.84 Lakhs

Less Depreciation = Rs 116012.53 Lakhs

Profit before Tax (PBT) = Sales Revenue –

Total Production Cost

= Rs 116012.53 Lakhs –

Rs 112962.59 Lakhs

= Rs 3049.94 Lakhs

Tax (33% Profit) = Rs 1006.48 Lakhs

Profit after Tax (PAT) = Rs 2043.46Lakhs



72

Cash accruals = Rs 2043.46 Lakhs

+ Rs 31.31 Lakhs

= Rs 2074.77 Lakhs

8.9 PAY BACK PERIOD

Pay Back Period (PBP) = Total Investment / Cash accurals

= 4094.41 Lakhs / Rs 2074.77 Lakhs

Pay Back Period (PBP) = 1.97 Years

8.10 RATE OF RETURN

Rate of Return = (Net Profit/total capital Investment)×100

= (2043.46/4094.41) × 100

= 49.9 %

\



73

9. PLANT LOCATION AND LAYOUT

Plant layout can be defined as the physical location or configuration of departments,

workstations and equipment in the production process. It is well defined arrangements of

physical resources which are used to produce the product. Plant layout is placing the right

equipment, coupled with the right method, in the right place to permit the processing of

product in the most effective manner through the shortest possible distance and in the shortest

possible time.

9.1 FACTORS AFFECTING THE LAYOUT

The following factors are to be considered.

Organization

Location of departments.

Type of product, method of production, production process, capacity.

Grouping of machines, material flow pattern.

Space requirement for the machines, work area, material handling, storage and other

facilities.

Safety factors, Health and other factors like ventilation, natural light removal of smoke,

and fumes etc.,

Provision for future expansion.

9.2 OBJECTIVE OF THE PLANT LAYOUT

Minimum material handling.



74

Facilitate manufacturing process.

Flexibility of arrangement.

Maximum utilization of equipment and floor area.

Care for employ safety and conveyance.

Easy supervision, minimum production delays, better working conditions.

9.3 ADVANTAGES OF SCIENTIFIC LAYOUT

It reduces internal transport to minimum.

It minimizes accidents and makes supervision easy and quick.

It makes repair and maintenance easy, reduces labour turnover and also reduces

production delays to a large extent.

It helps in production control and eliminates waste effort and thus

Increases the speed of production.

Possible to improve the production methods.

9.4 MAJOR CONSIDERATION FOR LAYOUT

Following are the major considerations which should be followed.

Various production departments such as Machine shop, welding shop and assembly

shop etc., should be located in the building. For this purpose product analysis should be made

which contains what is to be produced, how much is to be produced, what will be the sequence

of operations and how the machines can be best grouped.

The various production departments should be located in such a way that the traffic or

movement of workers should be easy. Flow of materials with in the department and also

between one department and another department should be economically and neatly managed.



75

Service department means canteen, stores, water etc., should be properly positioned

and should be suitably allotted. These should be located from factory as well as workers point

of view.

The various sections of departments of the factory can be arranged in to three main

groups, viz. Administrative, Services and Production departments. As far as possible,

administrative departments should be located in a separate wing, where there should be no

noise and disturbances due to movement and vibrations of machines. General Service

departments such as finished part store, tool rooms should be located as central as possible.

Storage facilities for raw materials and intermediate and finished products may be

located in isolated areas or in adjoining areas. Hazardous material becomes a decided menace

to life and property when stored in large quantities and should consequently be isolated.

Storage in adjoining areas to reduce materials handling may introduce an obstacle toward

future expansion of the plant. Arranging storage of materials so as to facilitate or simplify

handling is also a point to be considered in design. Where it is possible to pump a single

material to an elevation so that subsequent handling can be accomplished by gravity into

intermediate reaction and storage units, costs may be reduced. Liquids can be stored in small

containers, barrels, horizontal or vertical tanks and vats, either indoors or out of doors.

A great deal of planning is governed by local and national safety and fire code

requirements. Fire protection consisting of reservoirs, mains, hydrants, hoe houses, fire pumps,

reservoirs, sprinklers in buildings, explosion barriers and directional routing of explosion

forces to clear areas, and dikes for combustible-product storage tanks must be incorporated to

protect costly plant investment and reduce insurance rates.

Expansion must always be kept in mind. The question of multiplying the number of

units or increasing the size of the prevailing unit or units merits more study than it can be

given here. Suffice it to say that one must exercise engineering judgment; that has a penaltyfor bad judgment, scrapping of present serviceable equipment constitutes but one phase, for

shutdown due to remodeling may involve a greater loss of money than that due to rejected

equipment. Nevertheless, the cost of change must sometimes be borne, for the economies of

larger units may, in the end, make replacement imperative.



76

This word “Utilities” is now generally used for the ancillary services needed in the

operation of any production process. These services will normally be supplied from a central

site facility; and will include Electricity, Steam for process heating, Cooling water for general

use etc

The power required for electrochemical processes; motor drives, lighting, and general

uses, may be generated on site, but will more usually be purchased from the local supply

company. The voltage at which the supply is taken or generated will depend on demand. For a

large site the supply will be taken at a very high voltage, typically 11,000 or 33,000 V.

Transformers will be used to step down the supply voltage to the voltages used on the site. In

the United Kingdom a three-phase 415-V system is used for general industrial purposes, and

240-V single-phase for lighting and other low-power requirements. If a number of large

motors is used, a supply at an intermediate high voltage will also be provided, typically 6,000

or 11,000 V.

The steam for process heating is usually generated in water tube boilers; using the most

economical fuel available. The process temperatures required can usually be obtained with

low-pressure steam, typically 2.5 bar, and steam is distributed at a relatively low mains

pressure, typically around 8 bar. Higher steam pressures, or proprietary heat-transfer fluids,

such Dowtherm needed for high process temperatures. The generation, distribution and

utilization of steam for process heating in the manufacturing industries is properly planned.

The energy costs on a large site can be reduced if the electrical power required is

generated on site and the exhaust steam from the turbines used for process heating. The overall

thermal efficiency of such systems can be in the range of 70 to 80 % ; compared with the 30 to

40 % obtained from a conventional power station, where the heat in the exhaust steam is

wasted in the condenser. Whether a combined heat and power system scheme is worth

considering for a particular site will depend on the size of the site, the cost of fuel, the balance

between the power and heating demands; and particularly on the availability of, and cost of,

standby supplies and the price paid for any surplus power electricity generated.

Natural and forced-draft cooling towers are generally used to provide the cooling water

required on a site; unless water can be drawn from a convenient river or lake in sufficient



77

quantity. Sea water, or brackish water, can be used at coastal sites, but if used directly will

necessitate the use of more expensive materials of construction for heat exchangers.

The water required for general purposes on a site will usually be taken from the local

mains supply, unless a cheaper source of suitable quality water is available from a river, lake

or well.

Demineralised water, from which all the minerals have been removed by ion exchange,

is used where pure water is needed for process use, and a boiler feed water. Mixed and

multiple bed ion-exchange units are used; one resin converting the cations to hydrogen and the

other removing the acid radicals.

The distribution of gas, air, water, steam, power, and electricity is not always a major

item, inasmuch as the flexibility of distribution of these services permits designing to meet

almost any condition. But a little regard for the proper placement of each of these services,

practicing good design, aids in ease of operation, orderliness, and reduction in costs of

maintenance. No pipes should be laid on the floor or between the floor and the 7 feet level,

where the operator must pass or work. Chaotic arrangement of piping invites chaotic operation

of the plant. The flexibility of standard pipe fittings and power-transmission mechanisms

renders this problem one of minor difficult.

Existing or possible future railroads and highways adjacent to the plant must be known

in order to plan rail sidings and access roads within the plant. Railroad spurs and roadways of

the correct capacity and at the right location should be provided for in a traffic study and

overall master track and road plan of the plant area.

Some of the factors in rail-track planning are:

Existing and future off-site main rail facilities

Permissible radius of curvature for spurs- consult local rail authorities

Provision for traffic handling- arrangement of spurs and ladder track and switching



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Adequate spur facilities

Loading and unloading facilities for initial plant construction and subsequent

operations.

Rack stations for liquid handling.

Storage space for full and empty cars.

Space for cleaning and car repairs.

9.5 PLANT LAYOUT PROCEDURE

A number of procedures have been developed to facilitate the planning (Design) of

plant layouts. Following steps are generally considered in designing a plant layout. The

type of data and information required depends upon the nature of the layout problems. In case

planning a new plant layout, the data required will be very elaborate and will cover the

manufacturing particulars along with growth rate required. If the layout problem is for making

minor alterations in the layout, the data required will be limited to manufacturing methods,

sequence of operations, equipment involved etc. Product analysis consists of breaking down

the product into sub assemblies and the sub-assemblies into their individual parts. Product

analysis determines the operations necessary for the production of each part. Production

process is then analyzed.

Inter relationship between the work areas is also planned while designing the

individual operations or work areas. Similarly, the inter relationship between related groups of

operations should be worked out. At this stage, flow diagram also should be co coordinated

with the material flow pattern.

After installing the layout there may be possibility that it may not work as planned,

people may not follow the method they are supposed to, material does not arrive as planned. It

may result in delays. In such circumstances management should try to find out the reasons for

such delays, the activities which have to be taken care off and evaluate the consequences.

Action should be taken to rectify the mistakes, to minimize their effect on the execution of the

plan.



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9.6 PLANT LAYOUT

F ig. 9.1: Plant Layout



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10. MATERIAL HANDLING AND SAFETY

10.1 FACTS ABOUT H2S

It is highly toxic, colorless gas with rotten eggs smell.

It forms explosive mixture with air.

Prolonged exposure of low concentration H2S affects the organs responsible to

identified smell.

It dissolves in water and is very corrosive on carbon steel.

It severely attacks carbon steel above 3500C.

H2S detectors and alarms should be located at critical points.

Detectors should be capable of functioning during electrical power failure.

Personnel should wear breathing air masks whenever doing potentially hazardous

work. (e.g. Installing blind flanges, replacing orifice plates, opening sample or purge

connections, opening vessels for inspection or repair etc..)

During start-up and shut down, precautions must be taken against occurrence of

sulphur fires and explosions. Care must be taken to ensure that excess air is not

introduced during fuel gas firing mode (after first startup) as sulphur fires could result

when the temperature reaches 205oC.

H2S may be present anywhere in the plant, particularly in process areas. The most

common areas where H2S leaks may occur are

Pump seals

Fittings

Flanges

Drains



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Sample valves

Sulphur Rundown lines

10.2 HAZARD

H2S is generally recognized by a characteristic foul odor. Prolonged exposure to low

concentrations will dull the sense of smell. This can be fatal to those who think they can detect

a dangerous concentration by its offensive odor.

In low concentrations, H2S irritates the eyes, throat and respiratory system. When high

concentrations are present, death due to lung paralysis may occur before odor is detected.

Toxicity of H2S

1ppm – can smell

10 ppm – allowable for 8 hour exposure

100 ppm – kills smell in 3-5 min. May burn eyes and throat.

200 ppm – kills smell rapidly, burn eyes and throat. Will cause headache and nausea

300 ppm – Respiratory disturbances in 2-15 min. Artificial resuscitation needed.

Immediately dangerous to life and health.

500 ppm to 700 ppm – quick unconsciousness, breathing stops, immediate artificial

resuscitation needed.

1000 ppm – Unconsciousness at once, permanent brain damage unless rescued

promptly.



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10.3 FACTS ABOUT SULPHUR DIOXIDE (SO2)

It is produced when H2S burns

It is a very toxic gas and irritating gas

1 to 10 ppm – respiratory and pulse rate increases

20 ppm – It is highly irritating to eyes, nose, throat and lungs

It is equally toxic as compared to H2S and should be treated accordingly.

10.4 HAZARDS OF DIESEL

It is inflammable in nature; therefore it should not be subjected to very high

temperature.

Storage must be done in the vessels to withstand the increasing pressure because of its

volatility.

10.5 BASIC RESCUE PROCEDURE

Take the Victim to fresh air and apply artificial resuscitation.

Keep the Victim warm.

Do not give an unconscious person anything by mouth.

Do not leave the victim unattended even after he regains consciousness.

Take the victim to Doctor or call the doctor.

The most effective safety measure is the common sense approach.

Being aware of the hazard and act according



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10.6 ENVIRONMENT CONSIDERATIONS

Oil refineries are creating a major public health crisis because of the huge amounts of

toxic pollution they release on a daily basis. They release massive amount of Sulphur dioxide,

volatile organic compounds, particulate matter, nitrogen oxides and carbon monoxide, these

pollutants form ground level ozone and hazards. Thus, refineries also contribute to failure of

many areas to meet Environment Pollutants Assessment ambient air quality standards.

Refineries are the Nations major source of tons of toxic volatile organic compounds, like

cancer causing benzene, naphtha, ketones and chemical that cause asthma and childhood

development problems .Refineries are also a source of large chemical releases during fires

explosions, upsets and spills during the accidents. These spills often dump chemicals into

communities around refineries causing health problems. Most of toxic air pollution is formed

by product leaks in equipment, technology to prevent these leaks has been widely available.

Government regulations/norms are being adopted by Refineries to reduce pollution problems

so that better environment conditions are maintained.



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11. SUMMARY AND CONCLUSIONS

Hydroprocessing of renewable oil feed stocks creates a unique opportunity to produce

sustainable diesel fuel, completely compatible with existing fuel infrastructure, refinery

configuration and engine technology. The process has bright advantages in terms of energy

savings and cost, environmental friendliness, product quality and quantity and process

advantages; which are listed in this section.

11.1 ENVIRONMENTAL BENEFITS

The main advantage of the process is that it reduces the dependency on fossil fuel and

it is a clean fuel.

As the vegetable oils don’t have harmful emission precursors like N and S, it

contributes to the environment by reducing sulfur load in the atmosphere. It is found in

our studies that 5-8% sulfur is reduced in the product by co processing 10% Jatropha

oil along with diesel [sulfur content 2.3%] and it is consistent for at least 2 years of

operation.

The process doesn’t create the food chain implications as the oils are non edible. The

slight increase in CO2 release [approximately 0.1 -0.15 %] in the process will be well

compensated by the carbon consumption by the corresponding source plants/trees.

The energy savings and GHG (Green House Gases) saving in CO2 equivalents are

studied by CONCAWE (Conservation Of Clean Air And Water In Western Europe),

DOE [Department of Energy] and PNAS [Proceedings of National Academy of

Sciences, United States of America]. From this, it is understood that co-processing of

10% vegetable oil will save 4-4.3 GJ/ ton in comparison with that of blending 10%

Biodiesel [2.63 – 3.55 GJ/ ton] in the diesel.



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York, 1997.

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Publication, New York, 1999.

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