1

Bright Star University

Faculty of Engineering

Chemical Engineering Department

DESIGN OF OIL REFINERY TO OPERATE

200,000 BARREL PER DAY

A Project Submitted in Partial Fulfillment of the Requirements for

the Degree of B. Sc. In Chemical Engineering

By

Students student NO

Nawal Mohamed Ramadan 21152840

Qarnas Youssef Mohamed 21152849

Haniyeh Ibrahim Saleh 21152839

Anwar EL-Shamekh emragea 21152844

Supervisor

Dr. Raheel. G Raheel

Jun / 2019

2

CERTIFICATE

This project work entitled:

DESIGN OF OIL REFINERY TO OPERATE

200,000 BARREL PER DAY

Which is being submitted the following students

Nawal Mohamed Ramadan 21152840

Qarnas Youssef Mohamed 21152849

Haniyeh Ibrahim Saleh 21152839

Anwar EL-Shamekh emragea 21152844

In partial fulfillment of the requirements for the award of B.Sc. degree In

chemical engineering has been carried out under my supervisor and

accepted for representation.

Name ……………………………….…...

Signature ……………………….……….

Corrected approved copy for records

Name ……………………....... Name …………………….…...

Signature ……………………... Signature ………….………….

Date …………………….………

The members of the committee to examine the B.Sc. project find it

satisfactory and recommend that it be accepted.

1- Name Chair

2- Name Member

3- Name Member

Date

3

ABSTRACT

The worldwide oil refining industry currently faces strong challenges related to

concerns about future feedstock and characteristics of oil products. These challenges

favor two main strategies for the sector: the first strategy is increasing refinement

complexity and versatility; the second is integrating the refining and petrochemical

industries, adding value to the crude oil while guarantee market share to premium oil

products. Both strategies aim at increasing production of highly specified oil products,

simultaneously reducing the environmental impacts of the refining industry. The aim

of this study is design an oil refinery to operate a 200,000 barrel per day with a

feasibility building of this platform unit in Braga city.

Material balance, energy balance and a furnace design are done in this project,

Feasibility for the study is done, and also the unit operations are taken in the account

in this study.

4

ACKNOWLEDGMENT

بعد

شكر هللا عز وجل ال تعد وال تحصى التيعلى نعمه

نتقدمالهي ال يطيب الليل اإل بشكرك .. وال يطيب النهار اإل بطاعتك .. وال تطيب اللحظات اإل بذكرك .. وال تطيب اآلخرة اإل

بعفوك .. وال تطيب الجنة اإل برؤيتك .. إلى من بلغ الرسالة .. وأدى األمانة ونصح األمة .. إلى نبي الرحمة ونور

رسولنا محمد صلي هللا عليه وسلم...........................................................العالمين .........

تأخذمن إلىمن تسهر لننام وتجوع لنشبع إلىالقلب الحنون والحضن الدافئ إلىالتي تحترق لتضئ دربنا إلى الشمعة

إلىني فرحي ركمن تشا إلىاشعر به أنمن تحس بحزني قبل إلى أموريمن تعينني في كل إلىمن عمرها وتعطينا

أمي الحبيبة .....................................................................................................اعز البشر

إلى من جعله هللا سبباً في وجودي إلى مثلي األعلى ومعلمي في الحياة إلى قارب النجاة إلى الكرم والعطاء إلى الخير

أبي العزيز.....................................إلى من كان وراء كل نجاح أحققه أهدى هذا البحث حباً وتقديراً اليه ...

إلى سندي وقوة مالذي بعد هللا .. إلى من آثروني على نفسهم . إلى من علموني علم الحياة .. إلى من أظهروا لي ما

أخوتي وأخواتي....................................................................هو أجمل من الحياة .................

إلى من سطرت معهم على جدار الزمن أجمل الذكريات إلى نور العين إلى نبض القلب الذي به أحيا ولن

األصدقاء واألحباب.................................................................................انساهم...............

أساتذتي الكرام...............................إلى من وضح رؤيتي وأنار بالعلم طريقي ........إلى من علمني حرفاً

وكلمة الذين كفروا السفلى بدم الشهداء قديما وحديثا من أجل أن تكون كلمة هللا هى العليا تالذى ارتوالي االرض

يأرضنا الحبيبة ووطنا الغال........................................................................................

إلى كل هؤالء نقدم لهم كل

►شكر وعرفانال◄

5

وتقدير شكر

يسرنا أن نتقدم بخالص شكرنا وامتناننا إلى من كانوا سببا

، في إتمام وصولنا إلى هذه المرحلة في مسيرتنا الدراسية

أساتذتنا في المعهد العالي لبناء القدرات التقنية وجميع العاملين

في األقسام العلمية واإلدارية ، ونخص بالذكر منهم :

ه والمعلم في متابعته لنا وتيسيره الذي كان مثاالً للمشرف و المَوجِّ

لمراحل العمل في هذا المشروع

وال ننسى أن نتوجه بشكرنا إلى كل من مد يد العون لنا إلنجاح هذا

منهم : صوباألخالعمل

وهللا ولي التوفيق

-طلبة المشروع :

6

TABLE OF CONTENTS

ABSTRACT………………………………………………………………….. I

ACKNOWLEDGMENT………………………………………………….. II

LIST OF FIGURES………………………………………………………... VI

LIST OF TABLES…………………………………………………………. VII

LIST OF SYMBOLS……………………………………………………… VIII

CHAPTER (1): Overview

1.1: Introduction………………………………………………………….……... 1

1.2: History……………………………………………………………….……... 3

1.3: PETROLEUM …………..…………………………………………………. 4

1.3.1: The Composition of Petroleum …….……………………………... 4

1.3.2: Petroleum in Chemistry ……...…………………………………... 6

1.4: CLASSIFICATION OF PETROLEUM …………...……………...………. 8

1.5: 1.5. API GRAVITY ……………………………………………………...... 8

1.6 FLASH POINT ………………………...…………………………………... 10

1.7: VISCOCITY ……………………………..…………………….…………..

1.8: CRUDE DISTILLATION UNIT (CDU) ………………………...…………

1.9: REFLUX ……………………………………………………………………

10

11

15

CHAPTER (2): Unit Operations

2.1: Traditional Classification Of Processes……………………………………. 16

2.2: Essential Processes Based On Different Physical Principles………………. 16

2.3: Examples Of Unit Operations………………………………………............ 17

2.4: General Overview About Unit Operations………………………………….

18

CHAPTER (3): Material Balance

3.1: Introduction… …………………………………………………………….... 40

3.2: Mass Feedback (Recycle) ………………….………………………………. 41

3.3: Commercial Use……………………………………………………………. 41

3.4: Basic Calculation……………………………………… ………………….. 42

3.5: The Material Balance on The Stripper Towers …………………………….. 49

3.6: The Material Balances About The Stripper Of Kerosene ….……………….

3.7: Material Balance About The Stripper Of Diesel……………………………. 50

52

CHAPTER (4): Energy Balance

4.1: Types of Energy……………………………………… …………................ 54

4.2: Calculation Of Energy Balance For Heat exchanger……………………….. 55

7

CHAPTER (5): Design Of Heat Exchanger

5.1: SHELL AND TUBE HEAT EXCHANGER DESIGN…………..………... 63

5.1.1: Selection of tube material ………………………………………… 64

5.1.2: Applications and uses …………………………………………...…

5.2: SHELL AND TUBE EXCHANGERS: CONSTRUCTION DETAILS…… 65

66

5.3: STEPS FOR DESIGN OF HEAT EXCHANGER………………….……...

5.4: MODIFIED DESIGN………………………………………………………

5.5: OPTIMIZATION USING A CAD PROGRAM……………………………

5.6: SHAPE PARTS OF THE HEAT EXCHANGER………………………….

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78

82

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CHAPTER (6): Economic Study

6.1: FEASIBILITY STUDY……………………………..………………….….. 86

6.2: PLANT COST………………………….……………………………….….. 87

6.3: COST INDEX…………………………..………………………….............. 87

6.4: THE COST OF PRODUCTS AND RAW MATERIAL………….……….. 88

6.5: DEPRECIATION……………………………………………….………….. 90

6.6: THE COST OF THE OPERATION………...……………………………… 90

6.7: CASH FLOW……………………………………………………….............. 91

6.8: PAY- BACK PERIOD…………………………………………….………... 92

CONCLUSION……………………………………………………………….

94

RECOMMENDATION…………………………………………………….

94

REFERENCES……………………………………………………………….

APPENDIX

95

Appendix (A)…………………………………………………………………… 97

Appendix (B) …………………………………………………………………... 101

8

LIST OF FIGURES

Fig. 1.1: distillation tower ……………………………….............................. 12

Fig. 1.2: petroleum distillation unit …………………..…………………...

14

Fig. 2.1: Shell and tube Heat Exchanger………………………………..... 19

Fig. 2.2: Straight Tube Heat Exchanger (One Pass Tube-Side) …...… 20

Fig. 2.3: Straight Tube Heat Exchanger (Two Pass Tube Side) …...… 20

Fig. 2.4:Air-cooled exchangers bundle position……………………….... 21

Fig. 2.5: Fan cooler, Induced draft……………………………................... 21

Fig. 2.6: Fan cooler, Forced draft……………………………..................... 22

Fig. 2.7: Indirectly fired furnace……………………………...................... 23

Fig. 2.8: Cabin furnaces with horizontal tubes………………………...... 24

Fig. 2.9: Vertical cylindrical……………………………............................ 25

Fig. 2.10: Steam striping…………………………….................................. 27

Fig. 2.11: Stabilizer unit…………………………….................................. 29

Fig. 2.12: Separators three phase Horizontal Separator…………….…. 30

Fig. 2.13: Separators three phase Vertical Separator………………...… 31

Fig. 2.14: Separators three phase, Spherical Separator………………… 32

Fig. 2.15: Typical fixed-bed down flow catalytic reformer……...…… 34

Fig. 2.16: Reciprocating piston compressor………………………...…… 35

Fig. 2.17: Centrifugal compressor………………………………................. 37

Fig. 2.18: Centrifugal pump and Fluid path through the pump…...….. 38

Fig. 2.19: The hydraulic components in a single-stage in-line pump.. 38

Fig. 2.20: Schematics of ejectors………..…………………….................... 39

Fig. 3.1: percentages the product according the boiling point………... 44

Fig. 3.2: shows the logarithmic curve ……..……………………...............

Fig. 3.3: Material balance Stripper (H.N) ……..…………………............

Fig. 3.4: Material balance stripper (kerosene) ……..……………………

Fig. 3.5: Material balance strippers (diesel) ……..……………………....

Fig. 3.6: Diagram shows the unity of the oil refining………………..

46

50

51

53

53

Fig. 4.1: shows increase temperature crude oil inside heat exchanger 61

Fig. 5.1: U-tube heat exchanger ……………………………………........... 63

Fig. 5.2: Straight tube heat exchanger (one pass tube)…..…………….. 64

Fig. 5.3: Straight tube heat e xchanger (one pass 64

9

tube)…...…………….

Fig. 5.4U-tube heat exchanger constructions…………….……………...

Fig. 5.5: Design procedures for shell-and-tube heat exchangers……..

66

67

Fig. 5.6: temperature correction factors ……….........................................

Fig. 5.7: Shell-bundle clearances ……………….........................................

Fig. 5.8: Tube-side heat-transfer factor…………………………………...

Fig. 5.9: Tube-side friction factors ……………………………..…………

71

73

74

77

LIST OF TABLES

Tab. 1.1: chemical elements of petroleum …………...……..……….… 5

Tab. 1.2: the percentage of hydrocarbons in petroleum ………............

Tab. 1.3: Classification of Petroleum………………………………..

Tab. 1.4: Destinations for Straight-Run Distillates………………….

5

7

13

Tab. 2.1: Main types of pumps……………………………….................... 36

Tab. 3.1: range boiling temperature to each product …….……………. 43

Tab. 3.2: State work column……………………….. ………………… 43

Tab. 3.3: the percent of product with specific gravity …………...……. 46

Tab. 4.1: initial and final temperature for oil products …….……......... 55

Tab. 4.2: temperature crude oil in each unit…………………..……….... 62

Tab. 4.3: the overall energy balance ……………………………………...

Tab. 5.1. Shows Physical Properties for oil and kerosene……………

Tab. 5.2. Constant for use in equation tube arrangement……………..

62

69

72

Tab. 6.1: The flow rate of products and their prices….…………….… 89

Tab. 6.2: The flow rate of the raw material and their prices.………... 89

10

LIST OF SYMBOLS

SGoil Specific gravity of the crude oil

API American petroleum institute

Bbl Barrel

ρ Density

Ds Shell inside diameter

ut Tube-side velocity

Ao Heat transfer area

Ds Baffle spacing

Re Tube-side heat transfer coefficient

RWM Raw material

I.B.P Initial boiling point

F.B.P Final boiling point

Q The heat liberated

LPG Liquid petroleum gas

∆T temperature difference

NTubes Number of tubes

L.N Light naphtha

H.N Heavy naphtha

Gal Gallon

Re Reynolds number

jh Heat transfer factor

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12

CHAPTER ONE

INTRODUCTION

Oil refinery or petroleum refinery is an industrial process plant where crude oil is

transformed and refined into more useful products such as petroleum naphtha, gasoline, diesel

fuel, asphalt base, heating oil, kerosene, liquefied petroleum gas, jet fuel and fuel oils. Oil

refineries are typically large, sprawling industrial complexes with extensive piping running

throughout, carrying streams of fluids between large chemical processing units, such as

distillation columns. In many ways, oil refineries use much of the technology of, and can be

thought of, as types of chemical plants.The crude oil feedstock has typically been processed

by an oil production plant. There is usually an oil depot at or near an oil refinery for the

storage of incoming crude oil feedstock as well as bulk liquid products.Petroleum refineries

are very large industrial complexes that involve many different processing units and auxiliary

facilities such as utility units and storage tanks. Each refinery has its own unique arrangement

and combination of refining processes largely determined by the refinery location, desired

products and economic considerations.An oil refinery is considered an essential part of the

downstream side of the petroleum industry; some modern petroleum refineries process as

much as 800,000 to 900,000 barrels (127,000 to 143,000 cubic meters) per day of crude

oil.According to the Oil and Gas Journal in the world a total of 636 refineries were operated

on the 31 December 2014 for a total capacity of 87.75 million barrels (13,951,000

m3(.Jamnagar Refinery is the largest oil refinery. Located in Gujarat, India, it is owned by

Reliance (1)

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1.2. HISTORY

Prior to the nineteenth century, petroleum was known and utilized in various fashions

in Babylon, Egypt, China, Philippines, Rome and Azerbaijan. However, the modern history of

the petroleum industry is said to have begun in 1846 when Abraham Gessner of Nova

Scotia, Canada devised a process to produce kerosene from coal. Shortly thereafter, in

1854, Ignacy Lukasiewicz began producing kerosene from hand-dug oil wells near the town

of Krosno, Poland. The first large petroleum refinery was built in Ploesti, Romania in 1856

using the abundant oil available in Romania. In North America, the first oil well was drilled

in 1858 by James Miller Williams in Ontario, Canada. In the United States, the petroleum

industry began in 1859 when Edwin Drake found oil near Titusville, Pennsylvania. The

industry grew slowly in the 1800s, primarily producing kerosene for oil lamps. In the early

twentieth century, the introduction of the internal combustion engine and its use in

automobiles created a market for gasoline that was the impetus for fairly rapid growth of the

petroleum industry. The early finds of petroleum like those in Ontario and Pennsylvania were

soon outstripped by large oil "booms" in Oklahoma, Texas and California. Prior to World

War II in the early 1940s, most petroleum refineries in the United States consisted simply

of crude oil distillation units (often referred to as atmospheric crude oil distillation units).

Some refineries also had vacuum distillation units as well as thermal cracking units such

as visbreakers (viscosity breakers, units to lower the viscosity of the oil). All of the many

other refining processes discussed below were developed during the war or within a few years

after the war. They became commercially available within 5 to 10 years after the war ended

and the worldwide petroleum industry experienced very rapid growth. The driving force for

that growth in technology and in the number and size of refineries worldwide was the

growing demand for automotive gasoline and aircraft fuel.In the United States, for various

complex economic and political reasons, the construction of new refineries came to a virtual

stop in about the 1980s. However, many of the existing refineries in the United States have

revamped many of their units and/or constructed add-on units in order to: increase their

crude oil processing capacity, increase the octane rating of their product gasoline, lower

the sulfur content of their diesel fuel and home heating fuels to comply with environmental

regulations and comply with environmental air pollution and water pollution requirements.

(1)

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1.3. PETROLEUM

1.3.1. The Composition of Petroleum

In its strictest sense, petroleum includes only crude oil, but in common usage it

includes all liquid, gaseous and solid hydrocarbons. Under surface pressure and temperature

conditions, lighter hydrocarbons methane, ethane, propane and butane occur as gases, while

pentane and heavier hydrocarbons are in the form of liquids or solids. However, in an

underground oil reservoir the proportions of gas, liquid, and solid depend on subsurface

conditions and on the phase diagram of the petroleum mixture, oil well produces

predominantly crude oil, with some natural gas dissolved in it. Because the pressure is lower

at the surface than underground, some of the gas will come out of solution and be recovered

(or burned) as associated gas or solution gas. A gas well produces predominantly natural gas.

However, because the underground temperature and pressure are higher than at the surface,

the gas may contain heavier hydrocarbons such as pentane, hexane, and heptane in the

gaseous state. At surface conditions these will condense out of the gas to form "natural gas

condensate", often shortened to condensate. Condensate resembles gasoline in appearance and

is similar in composition to some volatile light crude oils.The proportion of light

hydrocarbons in the petroleum mixture varies greatly among different oil fields, ranging from

as much as 97 percent by weight in the lighter oils to as little as 50 percent in the heavier oils

and bitumens. (1).The hydrocarbons in crude oil are mostly alkanes, cycloalkanes and various

aromatic hydrocarbons, while the other organic compounds contain nitrogen, oxygen and

sulfur, and trace amounts of metals such as iron, nickel, copper and vanadium. Many oil

reservoirs contain live bacteria, the exact molecular composition of crude oil varies widely

from formation to formation but the proportion of chemical elements varies over fairly narrow

limits as follows:

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Table (1-1) chemical elements of petroleum

Element Percent range

Carbon 83 to 85%

Hydrogen 10 to 14%

Nitrogen 0.1 to 2%

Oxygen 0.05 to 1.5%

Sulfur 0.05 to 6.0%

Metals < 0.1%

Four different types of hydrocarbon molecules appear in crude oil. The relative

percentage of each varies from oil to oil, determining the properties of each oil. (1)

Table (1-2) the percentage of hydrocarbons in petroleum

Hydrocarbon Average Range

Alkanes (paraffins) 30% 15 to 60%

Naphthenes 49% 30 to 60%

Aromatics 15% 3 to 30%

Asphaltics 6% remainde

1.3.2. Petroleum in Chemistry

16

Petroleum is a mixture of a very large number of different hydrocarbons; the most

commonly found molecules are alkanes (paraffins), cycloalkanes (naphthenes), aromatic

hydrocarbons, or more complicated chemicals like asphaltenes. Each petroleum variety has a

unique mix of molecules, which define its physical and chemical properties, like color and

viscosity. The petroleum contains many of chemical compounds and principally contains of

the following compounds:

1- Paraffins: are also called alkanes and have the general formula of CnH2n+2, where n is the

number of carbon atoms. Paraffin's from C1 to C40 usually appeared in crude oil and represent

up to 20% of crude by volume. Since paraffin's are fully saturated (no double bond), they are

stable and remain unchanged over long periods of geological time, Paraffins are straight-chain

Paraffins or branched-chain Paraffins, For example:

CH3CH3

heptanen-

CH3 CH3

CH3

CH3

2,3-dimethylpentane

2- Aromatics: are an important series of hydrocarbons found in almost every petroleum

mixture from any part of the world. Aromatics are cyclic but unsaturated hydrocarbons that

begin with benzene molecule (C6H6) and contain carbon-carbon double bonds. Some of the

common aromatics found in petroleum and crude oils are benzene and its derivatives with

attached methyl, ethyl, propyl, or higher alkyl groups. This series of aromatics is called

alkylbenzenes and have a general formula of CnH2n-6 (where n ≥ 6). For example:

17

CH3 CH3

CH3

Benzene Toluene Xylene

3- Naphthenes or cycloalkanes are ring or cyclic saturated hydrocarbons with the general

formula of CnH2n. Cyclopentane (C5H10), cyclohexane (C6H12), and their derivatives such as

n-alkylcyclopentanes are normally found in crude oils. Thermodynamic studies show that

naphthenes rings with five and six carbon atoms are the most stable naphthenic hydrocarbons.

The content of Cycloparaffins in petroleum may vary up to 60%. For example:

R

Cyclopentane Cyclohexane Alkylcyclohexane

4- Olefins: are another series of noncyclic hydrocarbons but they are unsaturated and have at

least one double bond between carbon-carbon atoms. Compounds with one double bond are

called monoolefins or alkenes. monoolefins have a general formula of CnH2n. Olefins are

uncommon in crude oils due to their reactivity with hydrogen that makes them saturated;

however, they can be produced in refineries through cracking reactions. For example:

18

CH3 CH3

CH3 CH3

2,4-dimethylpentane

1.4. CLASSIFICATION OF PETROLEUM

Table (1-3) Classification of Petroleum (2).

Class of petroleum Composition of 250–300 °C fraction,

wt. %

Par. Napth Arom. Wax Asph.

Paraffinic 46—61 22–32 12–25 1.5–10 0–6

Paraffinic-naphtenic 42–45 38–39 16–20 1–6 0–6

Naphthenic 15–26 61–76 8–13 Trace 0–6

Paraffinic-naphtenic-aromatic

27–35 36–47 26–33 0.5–1 0–10

Aromatic 0–8 57–78 20–25 0–0.5 0–20

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1.5. API GRAVITY

API stands for the American Petroleum Institute, which is the major United

States trade association for the oil and natural gas industry. The API represents about 400

corporations in the petroleum industry and helps to set standards for production, refinement,

and distribution of petroleum products. They also advocate on behalf of the industry. One of

the most important standards that the API has set is the method used for measuring the

density of petroleum. This standard is called the API gravity.Specific gravity is a ratio of the

density of one substance to the density of a reference substance, usually water. The API

gravity is nothing more than the standard specific gravity used by the oil industry, which

compares the density of oil to that of water through a calculation designed to ensure

consistency in measurement. Less dense oil or “light oil” is preferable to more dense oil as it

contains greater quantities of hydrocarbons that can be converted to gasoline. Petroleum is

less dense that water and in 1916, the U.S. government instituted the Baumé scale as the

standard measure for any liquid less dense than water. This, in most cases, applies to oil. The

value used in this scale was 141.5 (see calculation below), but subsequent testing showed that,

due to error, the actual value should be 140. The government changed the scale to 140 to

correct the issue, but the use of 141.5 had become so entrenched in the oil industry that the

API decided to create the API gravity scale using the old value of 141.5.

API gravity is calculated using the specific gravity of oil, which is nothing more than

the ratio of its density to that of water (density of the oil/density of water). Specific gravity for

API calculations is always determined at 60 degrees Fahrenheit. API gravity is found as

follows:

API gravity = (141.5/Specific Gravity) – 131.5

Though API values do not have units, they are often referred to as degrees. So the

API gravity of West Texas Intermediate is said to be 39.6 degrees. API gravity moves

inversely to density, which means the denser oil is, the lower its API gravity will be. An API

of 10 is equivalent to water, which means any oil with an API above 10 will float on water

while any with an API below 10 will sink.

20

The API gravity is used to classify oils as light, medium, heavy, or extra heavy. As

the “weight” of oil is the largest determinant of its market value, API gravity is exceptionally

important. The API values for each “weight” are as follows:

Light – API > 31.1

Medium – API between 22.3 and 31.1

Heavy – API < 22.3

Extra Heavy – API < 10.0

These are only rough valuations as the exact demarcation in API gravity between

light and heavy oil changes depending on the region from which oil came. The fluctuation as

to what constitutes light crude in a given region is the result of commodity trading in oil.

Because density is a measure of weight per volume, API can be used to calculate how

many barrels of crude can be extracted from a metric ton of a given oil. A metric ton of West

Texas Intermediate, with an API of 39.6, will produce 7.6 barrels (at 42 gallons each). The

calculation is:

Barrels per metric ton = 1/ [(141.5/ (API + 131.5) x 0.159]. (1)

1.6. FLASH POINT

The flash point of an oil is the temperature at which the vapor above the oil will

momentarily flash or explode. This temperature is determined by laboratory testing using an

apparatus consisting of a closed cup containing the oil, heating and stirring equipment, and a

special adjustable flame. The type of apparatus used for middle distillate and fuel oils is called

the Pensky Marten (PM), while the apparatus used in

thecaseofKeroseneandlighterdistillatesiscalledtheAbel.Referencetothesetests are given later in

this Handbook, and full details of the tests methods and procedures are given in ASTM

Standards Part 7, Petroleum products and Lubricants.

There are many empirical methods for determining flash points from the ASTM

distillation curve. One such correlation is given by the expression Flash point ◦F=0.77 (ASTM

5% ◦F−150◦F). (1)

21

1.7. VISCOCITY

The viscosity of oil is a measure of its resistance to internal flow and is an indication

of its lubricating qualities, in the oil industry it is usual to quote viscosities either in

centistokes (which is the unit for kinematic viscosity), and seconds Saybolt universal, seconds

Saybolt furol, or seconds Redwood. These units have been correlated and such correlations

can be found in most data books. In the laboratory, test data on viscosities is usually

determined at temperatures of 100◦F, 130◦F, or 210◦F. In the case of fuel oils temperatures of

122◦F and 210◦F are used.(1)

1.8. CRUDE DISTILLATION UNIT (CDU)

Crude oil distillation is more complicated than product distillation, in part because

crude oils contain water, salts, and suspended solids. To reduce corrosion, plugging, and

fouling in crude heaters and towers, and to prevent the poisoning of catalysts in downstream

units, these contaminants are removed by a process called desalting. The two most typical

methods of crude-oil desalting – chemical and electrostatic separation – use hot water to

dissolve the salts and collect suspended solids. In chemical desalting, water and surfactants

are added to the crude, heated to dissolve salts and other impurities, and then sent to a settling

tank where the water and oil separate. In electrostatic desalting, chemicals are replaced with a

strong electrostatic charge, which drives the separation of water from oil.

Modern crude distillation towers can process 200,000 barrels of oil per day. They can

be up to 150 feet (50 meters) tall and contain 20 to 40 fractionation trays spaced at regular

intervals. In some towers, the trays in the top section are replaced with structured packing.

Before reaching the tower, desalted oil goes through a network of pre-heat exchangers to a

fired heater, which brings the temperature up to about 650°F (343°C). If the oil gets much

hotter than this, it starts to crack and deposit carbon inside the pipes and equipment through

which it flows

The hot crude enters the distillation tower just above the bottom. Steam is added to

enhance separation; it does so largely by decreasing vapor pressure in the column. When hot

oil enters the tower, most of it vaporizes. Unvaporized heavy fuel oil and/or asphalt residue

drops to the bottom of the tower, where it is drawn off. The vapors rise through the distillation

trays, which contain perforations and bubble caps (Figure 1-1). Each tray permits vapors from

below to bubble through the cooler, condensed liquid on top of the trayThis provides

excellent vapor/liquid contacting. Condensed liquid flows down through a pipe to the hotter

tray below, where the higher temperature causes re-evaporation. A given molecule evaporates

and condenses many times before finally leaving the tower. Products are collected from the

22

top, bottom and side of the column. Sidedraw products are taken from trays at which the

temperature corresponds to the cutpoint for a desired product. In modern towers, a portion of

each sidedraw stream is returned to the tower to control tray temperatures and further enhance

separation. Part of the top product is also returned; this “reflux” flow plays a major role in

controlling temperature at the top of the tower

Fig (1-1) distillation tower

23

After leaving the tower, product streams go to holding tanks or directly to

downstream process units. As shown in a Table (1-3), products can include heavy fuel oil,

heating oil, kerosene, gasoline, and uncondensed gases. (1)

Table (1-4) Destinations for Straight-Run Distillates

Fraction Approx. oC

Boling Range

OF

Next Destination Ultimate

Product(s)

LPG -40 to 0 -40 to 31 Sweetener Propane fuel

Light Naphtha 39 to 85 80 to 185 Hydrotreater Gasoline

Heavy Naphtha 85 to 200 185 to 390 Cat. Reformer Gasoline,

Aromatic

Kerosene 170 to 270 340 to 515 Hydrotreater Jet fuel, No. 1

diesel

Gas Oil 180 to 340 350 to 650 Hydrotreater Heating Oil, No.

2 diesel

Vacuum Gas Oil

340 to 566

650 to 1050

FCC

Hydrotreater

Lube plant

Hydrocracker

Fuel oil, FCC

feed, Lube

basestock,

Gasoline, jet,

diesel

24

Residue

>540

>1000

Coker

Visbreaker

Asphalt Unit

Hydrotreater

cooker gas oil

Visbreaker gas

oil, Deasphalted

oil

25

Fig (1-2) petroleum distillation unit

26

1.9 REFLUX

In normal distillation columns, heat is added to the column from a reboiler and

removed in an overhead condenser. A part of the distillate condensed in overhead condenser

is returned to the column as reflux to aid fractionation. This approach is not feasible in crude

distillation because the overhead temperature is too low for recovery of heat. Also the vapor

and liquid flows in column increase markedly from bottom to top, requiring a very large-

diameter tower. To recover the maximum heat and have uniform vapor and liquid loads in the

column, intermediate refluxes are withdrawn, they exchange heat with incoming crude oil

before entering the furnace and are returned to the plate above in the column (1)

27

28

CHAPTER TWO

UNIT OPERATIONS

Chemical Engineering has developed from chemical technology and became

an independent branch of knowledge. It is a generalized method of description of

processes and equipment of chemical and food technologies. Chemical Engineering

relies on mathematics, physics, chemistry and physical chemistry. The unit operations

are very important part in the chemical industries.

2.1. TRADITIONAL CLASSIFICATION OF PROCESSES

Mechanical processes

Hydro dynamical processes

Heat transport processes

Mass transport processes (or diffusion separation processes)

Chemical processes

Biochemical and biological processes

2.2. ESSENTIAL PROCESSES BASED ON DIFFERENT

PHYSICAL PRINCIPLES:

1) Transport Processes

Momentum transport – used in hydrodynamics and fluid flow.

Energy transport – used in description of heat transfer and heat exchangers.

Mass transport – used in diffusion-separation processes.

2) Transformation Processes

Phase change: it is used in heat exchange accompanied by boiling or

condensation and in mass change such as distillation, drying.

Chemical reaction: it is used in description of chemical reactors.

Unit Operations: it involve certain type of processes carried out in a specific

equipment .(2)

29

2.3. EXAMPLES OF UNIT OPERATIONS:

Fluid transportation, filtration settling, fluidization, mixing : they involve

hydro mechanical processes

Heat exchange, evaporation : they involve energy (heat) transport

Extraction, absorption, drying, and distillation, membranes: they involve mass

transport.

Reactors, bioreactors: they involve chemical reaction

The refinery is thus a very complex technical facility and its productivity can

be ensured only by strict and constant control of all the component parts, to guarantee

product quality, operational safety and environment protection, theses aims are

attained by measurement, control and supervision techniques, which are now of vital

importance in all petroleum and petrochemical complexes.Since there are numerous

and varied pieces of equipment involved they have been grouped in type categories

having similar functions. In this connection, the classification of chemical engineering

unit operations has been used and the following will be examined successively:

1. Equipment for separating the components of a mixture, either by mass

transfer between phases (gas-liquid or liquid-liquid) or by physical separation

of multiphase systems such as gas-liquid, liquid-liquid, gas-solid, and liquid-

solid mixtures.

2. Equipment for achieving heat transfers, such as furnaces and heat exchangers.

3. Equipment called reactors where chemical reactions take place.

4. Equipment used to accomplish the mechanical operations of fluid transport

( pumps, compressors, or mixing).( 2)

This chapter studies the types of the unit operations that is used in reformer and

hydrotreating unit

30

2.4. GENERAL OVERVIEW ABOUT UNIT OPERATIONS

1. Heat Exchangers

The purpose of heat exchanger:

a) - Cooling:

The gasoline product that is going to storage to reduce its Vapor pressure

General unit intercoolers remove the heat of reaction between reactors

Fractionator’s condensers condense the overhead, part of which may be the

product and the other part of which may be reflux that is returned to the

column to help effect a separation.

b) - Heating:

Fractionator’s rebuilders are used to add heat to fractionation column that

effects a separation.

Reactor charge heaters are used to heat the charge up to the reaction

temperature.

c) - To exchange heat between hot and cold process streams:

Feed exchanger that is used to heat the reactor charge by exchanging heat with

the reactor effluent.

Fractionator’s feed-bottoms exchanger that is used to heat the feed by

exchanging heat with the bottom.

Shell and Tube Heat Exchangers

A shell and tube heat exchanger is a class of heat exchanger designs. It is the

most common type of heat exchanger in oil refineries and other large chemical

processes and is suited for higher-pressure applications. It consists of a tube bundle

enclosed in a cylindrical casing called a shell. One fluid runs through the tubes, and

another fluid flows over the tubes (through the shell) to transfer heat between the two

fluids.Two fluids, of different starting temperatures, flow through the heat exchanger.

One flows through the tubes (the tube side) and the other flows outside the tubes but

inside the shell (the shell side). Heat is transferred from one fluid to the other through

31

the tube walls, either from tube side to shell side or vice versa. The fluids can be

either liquids or gases on either the shell or the tube side. In order to transfer heat

efficiently, a large heat transfer area should be used, so there are many tubes. In this

way, waste heat can be put to use. This is a great way to conserve energy. Typically,

the ends of each tube are connected to plenums through holes in tube sheets. The

tubes may be straight or bent in the shape of a U, called U-tubes. Most shell-and-tube

heat exchangers are 1, 2, or 4 pass designs on the tube side. This refers to the number

of times the fluid in the tubes passes through the fluid in the shell. In a single pass

heat exchanger, the fluid goes in one end of each tube and out the other.There are two

basic types of shell-and-tube exchangers. The first is the fixed tube sheet unit, in

which both tube sheets are fastened to the shell and the tube bundle is not removable.

The second type of shell-and-tube unit has one restrained tube sheet, called the

stationary tube sheet, located at the channel end. Differential expansion problems are

avoided by use of a freely riding floating tube sheet at the other end or the use of U

tubes.This design may be used for single or multiple pass exchangers. The tube

bundle is removable from the channel end, for maintenance and mechanical cleaning.

There are often baffles directing flow through the shell side so the fluid does not take

a short cut through the shell side leaving ineffective low flow volumes. Counter

current heat exchangers are most efficient because they allow the highest log mean

temperature difference between the hot and cold streams. Many companies however

do not use single pass heat exchangers because they can break easily in addition to

being more expensive to build. Often multiple heat exchangers can be used to

simulate the counter current flow of a single large exchanger. Shell-and-tube

exchangers are designed and fabricated according to the standards of the Tubular

Exchanger Manufacturers Association.(2)

32

Fig(2-1( Shell and tube Heat Exchanger

`

Fig(2-2) Straight Tube Heat Exchanger (One Pass Tube-Side: Countercurrent Flow)

33

Fig(2-3) Straight Tube Heat Exchanger (Two Pass Tube Side)

2. Fan Coolers

Air-cooled exchangers use atmospheric air as a cooling fluid. After being

sucked in by fans, the air goes through the usually finned tube bundles where a fluid

circulating inside is to be cooled or condensed. The bundles can be horizontal, vertical

or inclined, with the air and fluid on the tube side circulating in cross flow figure (2-

4), Bundles are horizontal in most air coolers employed with petroleum products.(2)

Fig(2-4) Air-cooled exchangers bundle position

A. Horizontal. B. Vertical. C. Inclined.

Induced draft: when the fans are placed above the bundles.

34

Fig(2-5) Fan cooler, Induced draft

Forced draft: when the fans are placed below the bundles.

Fig(2-6) Fan cooler, Forced draft

3. Heaters

Heater or Furnaces are equipment where fluids are heated by the gases

produced by combustion of a liquid or gaseous fuel. They are termed direct fired

35

furnaces, as the heat of the combustion gases is directly transmitted to the cold fluid

that circulates in a tubular coil. These furnaces are different from indirectly fired

furnaces where the process fluid to be heated circulates in a tube bundle submerged in

a hot fluid, which is itself heated directly by the fuel combustion gases figure (2-

7).(2).Direct fired furnaces must also be distinguished from mixing type furnaces

(such as catalyst regenerators) where the process fluid is raised to the specified

temperature by mixing with the fuel combustion gases.

Fig(2-7) Indirectly fired furnace

There are numerous layouts for tubes in the radiation and convection sections

and for one section in relation to the other. There are consequently many types of

furnaces, that can however be classified into the different categories:

Cabin Furnaces With Horizontal Tubes

36

Here the general shape of the radiation section is a parallelepiped, with the

greatest length being horizontal. The tubes are placed horizontally on the longest side

walls. The burners are located on the floor, on the lower part of the longest side walls

or on the end walls where there are no tubes. Figure (2-8).(2)

Figure (2-8) Cabin furnaces with horizontal tubes

Vertical Cylindrical

37

The radiation section is in the shape of a cylinder with vertical axis. The

burners are located on the floor at the base of the cylinder. The heat exchanger area

covers the vertical walls and therefore exhibits circular symmetry with respect to the

heating assembly figure (2-9). (2)

38

Fig(2-9) Vertical cylindrical

39

4. Steam Striping (Stripper)

A wastewater stream is heated and put in intimate contact with steam in a

packed or tray tower. The combined effects of the steam and heat, or temperature

cause organic material to transfer from the liquid to the vapor phase. This material is

then carried out with the vapor. As contacting proceeds down the tower, the

wastewater becomes leaner in the organic material while the vapor phase becomes

more enriched as it travels up the tower Steam is injected at the bottom of the tower to

provide heat and vapor flow. Clean water leaves the bottom of the tower. The

wastewater is fed at the top of the tower and the steam leaves the top heavily laden

with organic material. This steam/organic combination is condensed and processed

further as detailed in the next few pages. The net effect achieved in the steam stripper

and condenser is that a contaminated wastewater and steam are injected into the tower

and a clean water stream is obtained. Some organic materials are not totally miscible

in water and separate into a distinct organic phase when the concentration exceeds the

solubility limit. Most aromatics and halogenated organics fall in this category. Steam

stripping applications for these types of compounds can be very effective, since a

good part of the concentration of the organic can be accomplished in a decanter as

indicated in figure (3-10). In this case, the water layer is recycled to the stripping

column for reprocessing. The design of the decanter poses some interesting questions

since the water flow is generally significantly larger than the organic flow.

Furthermore, in some cases (benzene, toluene, etc), the organic layer is the lighter of

the two liquid phases. In applications involving halogenated organics, the organic

liquid is heavier than water. Needless to say, good models to predict the phase

behavior of the system in question are essential.(3)

40

Fig(2-10) Steam striping using structured packing. Water (light phase) is recycled as

reflux at top of stripper

41

5. Stabilizer

Condensate stabilization units are an important tool in making condensate

from natural gas streams easier to manage. Designs vary, but the goals of

implementing a stabilization unit are usually the same: increase recovery of

hydrocarbons, remove corrosive components, and create transport-ready product.

Carbon dioxide and hydrogen sulfide corrode transport infrastructure, and lighter

hydrocarbons like methane in the condensate are more dangerous to handle.

Removing these while optimizing recovery of constituents like methane and propane

means safer product and more profitable operations Raw natural gas liquids (NGLs)

are introduced into the stabilization unit at a high pressure through a shutdown valve

and preheated to a specific temperature. Afterwards, the pressure of the liquid feed is

dropped via a control valve, creating both a liquid and gas phase. At this point water

can be flashed out and removed from the stream, while any light gasses that escape

can be flashed off or recovered. The remaining hydrocarbon liquids are then heated

through a bottom exchanger before entering into a contactor or stabilizer tower.

Tower designs can vary from tray to packed column, but the idea is generally the

same: a reboiler heats the hydrocarbon liquid, causing two separate phases again.

Lighter hydrocarbons with lower boiling points rise up the tower as a gas, while the

heavier hydrocarbons contact the trays or packing in the column, collect, and stream

down to the bottom of the tower as a liquid. Using either flash drum liquids or another

technique, the NGLs are cooled and routed to a pipeline or storage facility. The vapor

that rose through the tower may be flashed off in a downstream flare (if permitted by

regulations) or routed to an NGL recovery skid designed to compress and store the

lighter hydrocarbons for later processing or use.Modern designs of condensate

stabilizer units have several advantages. They are increasingly designed to be of a

pretested, skid-mounted, modular design for rapid installation and start-up. These

modular units can be ordered based on specific needs, from low BPD well-site

recovery to high BPD midstream NGL processing. Hot or heat medium oil systems

help reduces emissions, while compression cycles can be optimized with digital

control systems. Additional equipment can be added or modified for special-case

scenarios as well.

42

Fig(2-11) Stabilizer unit

Stabilizer Unit Working Description

After upstream separation, the condensate stream enters the unit. It is first

preheated in the inlet heat exchanger and then flows into the upper section of the

stabilizer tower. Inside the tower, the light components – typically methane, ethane,

propane, and butane – rise and exit the overhead vapor outlet. The heavier liquid

components descend and pass through the indirect heater to cause additional vapor to

flash off, rise and exit the tower. Stabilized liquids collect in the bottom of the tower.

They are routed to the inlet heat exchanger to cool. Stabilized liquid product then

flows to storage tanks. The vapor that exits the tower flows through an ambient cooler

and then to a low-pressure separator. Gas from the separator is routed to compression

for delivery to market. NGLs exit the separator and flow to a blow case where they

are pressurized for a bullet tank or pipeline.(4)

43

6. Separators

A separator is a device used for separating oil, gases and water contained in

the effluent, this called be three phase separator, while a separator that used for

separating two phase called be two phase separator.

Three Phase Separators

1. Horizontal Separator

They have a very high exchange surface. These separators generally have a

smaller diameter than vertical separators for the same, quantity of gas and offer a

larger gas/liquid interface. They are also easier to mount on skids.

Fig(2-12) Separators three phase, Horizontal Separator

Advantages of horizontal separator

Horizontal separators are ideally suited to well streams having high gas-oil

ratios, constant flow, and small liquid surge. Horizontal separators are smaller and

less expensive than vertical separators for a given gas capacity. Liquid particles in the

wells stream travel horizontally and downward at the same time as a result of two

forces acting upon them-the horizontal force of the gas stream and the downward

force of gravity. Therefore, higher gas velocities can be permitted in horizontal

separators and still obtain the same degree of separation as in vertical separators.

Also, the horizontal separators have a much greater gas-liquid interface area than

other types which aids in the release of solution gas and reduction of foam, the

horizontal configuration is best suited for liquid-liquid-gas.

44

Disadvantages of horizontal separator

Becomes expensive and heavy when large sizes are required, liquid level

control is critical for horizontal separators; they are much harder to clean.

2. Vertical Separator

Three phase separator allowing separation of three phases oil, gas and water .

Fig(2-13) Separators three phase Vertical Separator

Advantages of vertical separator

The main advantage of this separator is that it can accept larger liquid slugs

without causing excessive entrainment in the gas. Considering the great distance there

usually is between the liquid level and the gas outlet, this reduces the entrainment of

liquid droplets. However, it has a larger diameter for any given gas capacity. Vertical

separators are therefore well suited to large quantities of liquid or, on the other hand

when there is only gas (the minimum liquid space in a horizontal drum is too great).

Typical applications are scrubbers, compressor suction, heating gas drums and certain

oil separators containing sediments. They are also used for wells with solid deposits

(easy to clean).

45

Disadvantages of vertical separator

Equipment is expensive and heavy when large sizes are required. A vertical

separator for the same capacity is usually larger than a horizontal separator.

3. Spherical Separator

These types of separators are not very common and are reserved for wells with

very high gas-oil ratios (gas + condensate wells).

Fig(2-14) Separators three phase, Spherical Separator

Advantages of spherical separator

They are relatively low-cost and compact, but with a limited liquid retention

time and decantation section. Their advantage is their compact size but they offer

small capacities. They are very easy to handle. Another interesting point of

importance is that owing to their spherical shape, they can support all pressure ranges

Disadvantages of spherical separator

Use in three-phase separation is, therefore, very difficult if not impossible.

Two Phase Separators

This type of operation is very frequently encountered in actual practice in

most processes. Gravity separation is then often utilized due to its simplicity and low

46

cost. The problem of separating either a vapor phase and a liquid phase or two

immiscible liquid phases arises

constantly in the refinery. The operation is carried out in vertical or horizontal drums

whose seizing is such that two objectives are accomplished: residence time is

sufficiently long for the separation to occur as completely as possible and

entrainments of one in the other are reduced to a minimum.(5)

Gas-Liquid separation

Liquid-Liquid separation

Gas-Solid separation

Liquid-Solid separation

7. Distillation Tower

An apparatus used widely for countercurrent contacting of vapor and liquid to

effectseparations by distillation or absorption. In general, the apparatus consists of a c

ylindricalvessel with internals designed to obtain multiple contacting of ascending vap

or and descendingliquid, together with means for introducing or generating liquid at t

he top and vapor at thebottom.In a column that can be applied to distillation (see illust

ration), a vapor condenser is used toproduce liquid (reflux) which is returned to the to

p, and a liquid heater (reboiler) is used togenerate vapor for introduction at the bottom

. In a simple absorber, the absorption oil is the topliquid and the feed gas is the bottom

vapor. In all cases, changes in composition produce heateffects and volume changes,

so that there is a temperature gradient and a variation in vapor,and liquid flows from t

op to bottom of the column. These changes affect the internal flow ratesfrom point to

point throughout the column and must be considered in its design.Distillation columns

used in industrial plants range in diameter from a few inches to 40 ft (12 m)and in hei

ght from 10 to 200 ft (3 to 60 m). They operate at pressures as low as a fewmillimeter

s of mercury and as high as 3000 lb/in.2 (2 megapascals) at temperatures from 300to 7

00°F (180 to 370°C). They are made of steel and other metals, of ceramics and glass,

andeven of such materials as bonded carbon and plastics.A variety of internal devices

have been used to obtain more efficient contacting of vapor and liquid. The most

widely used devices are the bubble cap plates the perforated or sieve plate, and the

packed column.

47

Fig(2-15) Distillation Tower

The bubble cap plate is a horizontal deck with a large number of chimneys over which

circular or rectangular caps are mounted to channel and distribute the vapor through

the liquid. Liquid flows by gravity downward from plate to plate through separate

passages known as down comers. (5)

8. Compressors

A compressor is a machine designed to raise the pressure of compressible fluid

that passes through it. The name indicates that the fluid becomes compressed (its

volume decreases) as the pressure increases. Since gases are compressible fluids they

require compressors, while liquid require pumps as they are practically

incompressible. Gases can be considered incompressible for very low compression

48

ratios and blowers or fans are used. The compressors most commonly used in a

refinery are reciprocating piston compressors in the positive displacement category

and centrifugal compressors in the dynamic category.

Reciprocating Piston Compressor

This type of compressor used a proven technology that has been known for

some time and is the most widely employed of the positive displacement compressors

for power rating of 100 to 5000 KW.

Fig(2-16) Reciprocating piston compressor

Centrifugal Compressor

These compressors are widely used in refining and in the chemical and

petrochemical industries. They are built according to API 617 standards, are very

compact and can develop significant power in comparison with their size.( 2)

49

Fig(2-17) Centrifugal compressor

9. Pumps

Pumps allow liquids to be circulated and are found in different forms in

refining units. They can be classified into two main categories:

1- Centrifugal pumps.

2- Positive displacement pumps.

Table (2-1) Main types of pumps

Category Type Structure

Centrifugal

Single-stage

Multi-stage

Volute

Diffuser

Regenerative

Vertical

Helico-centrifugal

Axial flow

positive displacement

Rotary Gear-Lobe

Screw-Peripheral

Vane -Progressive cavity

Reciprocating Piston

Diaphragm

Plunger

50

There are a great many subgroups as indicated in table (2-1). Choosing

among the different types depends on numerous parameters, some of the most

important being the capacity (flow rate m3/h), the pressure head that is generated by

the pump, the type of liquid pumped (in particular its viscosity and vapor pressure

under inlet conditions). An initial selection is generally made on basis of the first two

criteria mentioned, i.e. the capacity and the pressure generated. The centrifugal pump

is often the only possible choice for high capacities, whereas positive displacement

pumps are better suited to generating high pressure differenc Other criteria such as the

viscosity of the pumped liquid can modify this initial choice. A positive displacement

pump is generally recommended to pump liquid with a viscosity higher than 2000cP.

In actual practice there are often situation where the choice requires taking other

parameters, in particular economic ones, into account.(5)

Centrifugal Pumps

In its simplest form, a centrifugal pump consists of impeller equipped with

radial vanes rotating inside a shell called the pump casing. The working principle is to

use the centrifugal force generated by the rotating impeller to transfer kinetic energy

to the pumped liquid This energy is then converted into pressure when the fluid

velocity decreases. Principle of the centrifugal pump An increase in the fluid pressure

from the pump inlet to its outlet is created when the pump is in operation. This

pressure difference drives the fluid through the system or plant.The centrifugal pump

creates an increase in pressure by transferring mechanical energy from the motor to

the fluid through the rotating impeller. The fluid flows from the inlet to the impeller

center and out along its blades. The centrifugal force hereby increases the fluid

velocity and consequently also the kinetic energy is transformed to pressure.

51

Fig(2-18) Centrifugal pump and Fluid path through the pump

Hydraulic components: The principles of the hydraulic components are

common for most centrifugal pumps. The hydraulic components are the parts in

contact with the fluid. Figure (2-19) shows the hydraulic components in a single-stage

in-line pump. (5)

Fig(2-19) The hydraulic components in a single-stage in-line pump.

52

Advantages of centrifugal pump

The centrifugal pump is the most used pump type in the world. The principle is

simple, well-described and thoroughly tested, and the pump is robust, effective and

relatively inexpensive to produce. There is a wide range of variations based on the

principle of the centrifugal pump and consisting of the same basic hydraulic parts.

Disadvantages of centrifugal pump

The centrifugal pump is not better suited to generating high pressure

difference, and there problem are that happen in the pump, the liquid specification

problem (Cavitation, Air pockets, Scales and Corrosion), operational problem and

mechanical problem.(6)

10. Ejectors

An ejector is a device that uses a motive fluid passing through a diffuser at

high speed. A negative pressure is thus generated that entrains another fluid. The

principle is in fact an application of the Bernoulli theorem that expresses the

conservation of momentum in a system. The steam expands is entropic ally through

the entrance orifice in the suction chamber. It then mixes with the gas being evacuated

at constant pressure and with no loss in momentum. A large part of the steam kinetic

energy is converted in to heat. The mixture then flows through the diffuser, with the

mixture kinetic energy being converted into a pressure increase. Isentropic

compression efficiency is approximately 80 %. (2)

Fig(2-20) Schematics of ejectors A. Single nozzle. B. Multiple nozzles

53

54

CHAPTER THREE

MATERIAL BALANCE

3.1. INTRODUCTION

Material balances are important first step when designing a new process or

analyzing an existing one. They are almost always prerequisite to all other calculations in the

solution of process engineering problems. Material balances are nothing more than the

application of the law of conservation of mass, which states that mass can neither be

created nor destroyed. Thus, you cannot, for example, specify an input to a reactor of one

ton of naphtha and an output of two tons of gasoline or gases or anything else. One ton of

total material input will only give one ton of total output, i.e. total mass of input = total mass

of output. A material balance is an accounting for material. Thus, material balances are often

compared to the balancing of current accounts. They are used in industry to calculate mass

flow rates of different streams entering or leaving chemical or physical processes.

The General Balance Equation Suppose propane is a component of both the input and

output streams of a continuous process unit shown below, these flow rates of the input and

output are measured and found to be different.

If there are no leaks and the measurements are correct, then the other possibilities

that can account for this difference are that propane is either being generated, consumed,

or accumulated within the unit.

A balance (or inventory) on a material in a system (a single process unit, a collection

of units, or an entire process) may be written in the following general way:

. Input + generation − output − consumption = accumulation

(Enter (produced (leave (consumed (buildup

Through within through within within

System) system) system) system) system)

55

This general balance equation may be written for any material that enters or

leaves any process system; it can be applied to the total mass of this material or to any

molecular or atomic species involved in the process. The general balance equation

may be simplified according to the process at hand. For example, by definition, the

accumulation term for steady-state continuous process is zero. Thus the above

equation becomes:

Input + generation = output + consumption

For physical process, since there is no chemical reaction, the generation and

consumption terms will become zero, and the balance equation for steady-state

physical process will be simply reduced to:

Input = Output

3.2. MASS FEEDBACK (RECYCLE)

Mass balances can be performed across systems which have cyclic flows. In

these systems output streams are fed back into the input of a unit, often for further

reprocessing.

Such systems are common in grinding circuits, where materials are crushed then

sieved to only allow a particular size of particle out of the circuit and the larger

particles are returned to the grinder. However, recycle flows are by no means

restricted to solid mechanics operations; they are used in liquid and gas flows, as well.

One such example is in cooling towers, where water is pumped through a tower many

times, with only a small quantity of water drawn off at each pass (to prevent solids

build up) until it has either evaporated or exited with the drawn off water.

The use of the recycle aids in increasing overall conversion of input products,

which is useful for low per-pass conversion processes (such as the Haber

process).(7)

3.3. COMMERCIAL USE

In industrial process plants, using the fact that the mass entering and leaving any

portion of a process plant must balance, data validation and reconciliation algorithms may

be employed to correct measured flows, provided that enough redundancy of flow

56

measurements exist to permit statistical reconciliation and exclusion of detectably

erroneous measurements. Since all real world measured values contain inherent error, the

reconciled measurements provide a better basis than the measured values do for financial

reporting, optimization, and regulatory reporting. Software packages exist to make this

commercially feasible on a daily basis .(7)

3.4. BASIC CALCULATIONS

The basic calculation for design the units of refinery is 200000 𝑏𝑎𝑟𝑙/𝑑𝑎𝑦

200000 𝑏𝑎𝑟𝑙 159 𝑙𝑖𝑡𝑒𝑟 1 𝑘𝑔

𝑑𝑎𝑦 1 𝑏𝑎𝑟𝑟𝑒𝑙 1 𝑙𝑖𝑡𝑒𝑟

= 31800000 𝑘𝑔/𝑑𝑎𝑦

The specific gravity of the crude oil from table specificationat temperature 15.7 ℃ is 0.8418

𝑆𝐺𝑜𝑖𝑙 =𝜌𝑜𝑖𝑙

𝜌𝐻2𝑜

𝜌𝑜𝑖𝑙 = 𝑆𝐺𝑜𝑖𝑙 × 𝜌𝐻2𝑂 = 0.8418 × 1000 𝐾𝑔

𝑚3

𝜌𝑜𝑖𝑙 = 841.8 𝐾𝑔

𝑚3

𝑉𝑜𝑖𝑙 = 𝑚𝑜𝑖𝑙

𝜌𝑜𝑖𝑙=

31800000𝐾𝑔/𝑑𝑎𝑦

841.8 𝐾𝑔/𝑚3

𝑉𝑜𝑖𝑙 = 37776𝑚3

𝑑𝑎𝑦

57

𝐴𝑃𝐼 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 =141.5

𝑆𝐺− 131.5

=141.5

0.8418− 131.5 = 36.5921

From table of distillation products (ASTM) for oil and from TBP chart (2.1),the volume

percent (V %) of all products the distillation column is calculated

Table (3-1) range boiling temperature to each product (ASTM)

Light

naphtha

Heavy

naphtha

Kerosene diesel residue

I.P.P

10

30

50

70

90

95

F.P.B

20

54

80

99

116

137

-

167

190

160

162

166

170

176

180

206

148

180

198

204

208

216

220

242

192

246

270

280

304

326

334

348

230

342

358

-

-

-

-

358

Table (3-2) State work column (temperature – pressure)

58

Heat outlet temp

Heavy Naphtha

Kerosene

Diesel

Residue

Flash zoon temp

Bottom temp

Top temp

Top pressure

Level

343 ℃

153 ℃

184 ℃

256 ℃

320 ℃

325 ℃

320 ℃

98 ℃

0.56 kg/cm3

52%

Figure (3-1) percentages the product according the boiling point

The percent of each product in the crude oil is calculated by using the relationship

betweentemperature range (I.B.P and FBP) and percent volume (𝑉%) fromtable (2.1)

andfigure (2.1) drawing by excel program.

59

Where the factor of steam that injects with the product increases the temperature of the

product in the equipment. The calculations are mentioned below:

LPG from 10 ℃𝑡𝑜→ 90 ℃

𝑉𝐿𝑃𝐺% = 𝑉𝑖 + 𝑉𝑓

2 × 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑡𝑒𝑎𝑚 (𝐿𝐵𝐺)

= 1.5 + 7

2= 4.25 × 0.33 = 1.4

𝐿. 𝑁 𝑓𝑟𝑜𝑚 20℃ → 167℃

𝑉𝐿. 𝑁% = 𝑉𝑖 + 𝑉𝐹

2 × 𝐹𝑎𝑐𝑡𝑜𝑟 𝑠𝑡𝑒𝑎𝑚 (𝐿. 𝑁)

𝑉𝐿. 𝑁% = 15.5 + 26

2 × 0.29 = 6

𝑉𝐻. 𝑁% = 𝑉𝑖 + 𝑉𝐹

2 × 𝐹𝑎𝑐𝑡𝑜𝑟 𝑠𝑡𝑒𝑎𝑚 (𝐻. 𝑁)

𝑉𝐻. 𝑁% = 2 + 20

2 × 0.72 = 15.1

𝐾𝑒𝑟 𝑓𝑟𝑜𝑚 148℃ → 242℃

𝑉𝐾𝐸𝑅 % = 𝑉𝑖 + 𝑉𝐹

2 × 𝐹𝑎𝑐𝑡𝑜𝑟 𝑠𝑡𝑒𝑎𝑚 (𝐾𝑒𝑟)

𝑉𝐾𝐸𝑅 % = 18 + 32

2 × 0.22 = 5.5

𝐷𝑖𝑠𝑒𝑙 𝑓𝑟𝑜𝑚 192℃ 𝑡𝑜→ 348℃

𝑉𝐷𝑖𝑒𝑠𝑒𝑙% = 𝑉𝑖 + 𝑉𝐹

2 × 𝐹𝑎𝑐𝑡𝑜𝑟 𝑠𝑡𝑒𝑎𝑚 (𝐷𝑖𝑠𝑒𝑙)

𝑉𝐷𝑖𝑒𝑠𝑒𝑙% = 25 + 52

2 × 0.31 = 12

𝑅𝑒𝑠𝑖𝑑𝑢𝑚 𝑓𝑟𝑜𝑚 230℃ 𝑡𝑜→ 358℃

= 𝑉𝑖 + 𝑉𝐹

2 × 𝐹𝑎𝑐𝑡𝑜𝑟 𝑠𝑡𝑒𝑎𝑚 (𝑅𝑒𝑠𝑖𝑑𝑢𝑚)

𝑉𝑅𝑒𝑠𝑖𝑑𝑢𝑚% = 30 + 52

2 × 1.46 = 59.86

60

After the calculation of the percent products, the volume of the flow rates for all products

could be found:

𝑉𝐿𝑃𝐺 = 𝑉𝑜𝑖𝑙 × 𝑝𝑒𝑟𝑐𝑒𝑛𝑡(𝐿𝑃𝐺)

= 37776𝑚3

𝑑𝑎𝑦 × 0.014 = 528.864

𝑚3

𝑑𝑎𝑦

𝑉𝐿.𝑁 = 𝑉𝑜𝑖𝑙 × 𝑝𝑒𝑟𝑐𝑒𝑛𝑡(𝐿. 𝑁)

= 37776𝑚3

𝑑𝑎𝑦 × 0.06 = 2266.56

𝑚3

𝑑𝑎𝑦

𝑉𝐻.𝑁 = 𝑉𝑜𝑖𝑙 × 𝑝𝑒𝑟𝑐𝑒𝑛𝑡(𝐻. 𝑁)

= 37776𝑚3

𝑑𝑎𝑦 × 0.151 = 5704.176

𝑚3

𝑑𝑎𝑦

𝑉𝐾𝑒𝑟 = 𝑉𝑜𝑖𝑙 × 𝑝𝑒𝑟𝑐𝑒𝑛𝑡(𝐾𝑒𝑟)

= 37776𝑚3

𝑑𝑎𝑦 × 0.055 = 2077.68

𝑚3

𝑑𝑎𝑦

𝑉𝐷𝑖𝑒𝑠𝑒𝑙 = 𝑉𝑜𝑖𝑙 × 𝑝𝑒𝑟𝑐𝑒𝑛𝑡(𝐷𝑖𝑒𝑠𝑒𝑙)

= 37776𝑚3

𝑑𝑎𝑦 × 0.12 = 4533.12

𝑚3

𝑑𝑎𝑦

𝑉𝑟𝑒𝑠𝑖𝑑𝑢𝑚 = 𝑉𝑜𝑖𝑙 × 𝑝𝑒𝑟𝑐𝑒𝑛𝑡(𝑟𝑒𝑠𝑖𝑑𝑢𝑚)

= 37776𝑚3

𝑑𝑎𝑦 × 0.5986 = 22612.7136

𝑚3

𝑑𝑎𝑦

𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑉𝐿𝑃𝐺 + 𝑉𝐿.𝑁 + 𝑉𝐻.𝑁 + 𝑉𝐾𝑒𝑟 + 𝑉𝐷𝑖𝑒𝑠𝑒𝑙 + 𝑉𝑟𝑒𝑠𝑖𝑑𝑢𝑚

𝑉𝑡𝑜𝑡𝑎𝑙 = 528.864 + 2266.56 + 5704.176 + 2077.68 + 4533.12 + 22612.7136

= 37723.1136𝑚3

𝑑𝑎𝑦

The calculation of the specific gravity of all products is done by using Figure (2.2) that

shows the relation between percent of volume of the product (V%) and the specific density as

following:

61

Figure(3-2) shows the logarithmic curve between the percent of the product and specific

weight

Table (3-3) the percent of product with specific gravity as resulting from figure( 2-2)

V% S.G

1.4

15.1

6

5.5

1.2

60

0.67

0.74

0.78

0.79

0.84

0.90

The calculation of the density for the product with the percent of the oil is done as following:

𝑆. 𝐺 = 𝜌𝑝𝑟𝑜𝑑𝑢𝑐𝑡

𝜌𝑤𝑎𝑡𝑒𝑟

62

𝜌𝐿𝑃𝐺 = 𝑆. 𝐺𝐿𝑃𝐺 × 𝜌𝑤𝑎𝑡𝑒𝑟 = 0.67 × 1000𝐾𝑔

𝑚3= 670

𝐾𝑔

𝑚3

𝑚𝐿𝑃𝐺 = 𝜌𝐿𝑃𝐺 × 𝑉𝐿𝑃𝐺 = 670 𝐾𝑔

𝑚3 × .528.864

𝑚3

𝑑𝑎𝑦= 354338.88

𝐾𝑔

𝑑𝑎𝑦

𝜌𝐿.𝑁 = 𝑆. 𝐺𝐿.𝑁 × 𝜌𝑤𝑎𝑡𝑒𝑟 = 0.74 × 1000 𝐾𝑔

𝑚3= 740

𝐾𝑔

𝑚3

𝑚𝐿.𝑁 = 𝜌𝐿.𝑁 × 𝑉𝐿.𝑁 = 740 𝐾𝑔

𝑚3 × 2266.56

𝑚3

𝑑𝑎𝑦= 1677254.4

𝐾𝑔

𝑑𝑎𝑦

𝜌𝐻.𝑁 = 𝑆. 𝐺𝐻.𝑁 × 𝜌𝑤𝑎𝑡𝑒𝑟 = 0.78 × 1000 𝐾𝑔

𝑚3= 780

𝐾𝑔

𝑚3

𝑚𝐻.𝑁 = 𝜌𝐻.𝑁 × 𝑉𝐻.𝑁 = 780 𝐾𝑔

𝑚3 × 5704.176

𝑚3

𝑑𝑎𝑦= 4449257.28

𝐾𝑔

𝑑𝑎𝑦

𝜌𝐾𝑒𝑟 = 𝑆. 𝐺𝐾𝑒𝑟 × 𝜌𝑤𝑎𝑡𝑒𝑟 = 0.79 × 1000 𝐾𝑔

𝑚3= 790

𝐾𝑔

𝑚3

𝑚𝐾𝑒𝑟 = 𝜌𝐾𝑒𝑟 × 𝑉𝐾𝑒𝑟 = 790 𝐾𝑔

𝑚3 × 2077.68

𝑚3

𝑑𝑎𝑦= 1641367.2

𝐾𝑔

𝑑𝑎𝑦

𝜌𝐷𝑖𝑒𝑠𝑙 = 𝑆. 𝐺𝐷𝑖𝑒𝑠𝑙 × 𝜌𝑤𝑎𝑡𝑒𝑟 = 0.84 × 1000 𝐾𝑔

𝑚3= 840

𝐾𝑔

𝑚3

𝑚𝐷𝑖𝑒𝑠𝑙 = 𝜌𝐷𝑖𝑒𝑠𝑙 × 𝑉𝐷𝑖𝑒𝑠𝑙 = 840 𝐾𝑔

𝑚3 × 4533.12

𝑚3

𝑑𝑎𝑦= 3807820.8

𝐾𝑔

𝑑𝑎𝑦

𝜌𝑟𝑒𝑠𝑖𝑑𝑢𝑚 = 𝑆. 𝐺𝑟𝑒𝑠𝑖𝑑𝑢𝑚 × 𝜌𝑤𝑎𝑡𝑒𝑟 = 0.90 × 1000 𝐾𝑔

𝑚3= 900

𝐾𝑔

𝑚3

𝑚𝑟𝑒𝑠𝑖𝑑𝑢𝑚 = 𝜌𝑟𝑒𝑠𝑖𝑑𝑢𝑚 × 𝑉𝑟𝑒𝑠𝑖𝑑𝑢𝑚 = 900 𝐾𝑔

𝑚3 × 22612.7136

𝑚3

𝑑𝑎𝑦= 20351442.24

𝐾𝑔

𝑑𝑎𝑦

63

The next calculations where done to make sure that the material balance are correct and found

percent error:

𝑚𝑡𝑜𝑡𝑎𝑙 = 528864 + 5704176 + 2266560 + 2077680 + 4533120 + 22612713.6

= 37723113.6 𝐾𝑔

𝑑𝑎𝑦

37723113.6 𝐾𝑔

𝑑𝑎𝑦 ×

1 𝑚3

1000 𝐾𝑔= 37723.1136

𝑚3

𝑑𝑎𝑦

𝑚 = 𝑉𝑜𝑖𝑙 × 𝜌𝑜𝑖𝑙 = 37723.1136𝑚3

𝑑𝑎𝑦 × 841.8

𝐾𝑔

𝑚3= 31755317.03

𝐾𝑔

𝑑𝑎𝑦

Where: 31755317.03 𝐾𝑔

𝑑𝑎𝑦 ×

1𝑙𝑖𝑡𝑒𝑟

1𝑘𝑔 ×

1𝑏𝑎𝑟𝑟𝑒𝑙

159 𝑙𝑖𝑡𝑒𝑟

Vtotal = 199718.975𝑏𝑎𝑟𝑟𝑒𝑙

𝑑𝑎𝑦

Product Percent

volume%

Volume

product 𝑚3

𝑑𝑎𝑦

S.G 𝑑𝑒𝑛𝑠𝑖𝑡𝑦

𝐾𝑔

𝑚3 𝑚𝑎𝑠𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

𝐾𝑔

𝑑𝑎𝑦

M.M

LPG

1.4

528.864 0.67

-

670

354338.88

L.N

15.1 5704.176 0.734

740

4449257.28 100

H.N

6

2266.56 0.78

780

1677254.4 150

Kerosene

5.5

2077.68 0.79

790

1641367.2 165

Diesel

12

4533.12 0.84

840

3807820.8 215

Residue 60 22612.7136 0.90 900 20351442.24 320

64

𝑒𝑟𝑟𝑜𝑟 = 𝑡𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝑓𝑎𝑙𝑠𝑒 𝑣𝑎𝑙𝑢𝑒

𝑡𝑟𝑢𝑒 𝑓𝑎𝑙𝑢𝑒

= 200000 − 199718.975

200000 × 100 = 0.14%

Out put In put

200000 199718.975

3.5. The Material Balance on The Stripper Towers:

The calculations of the inputs and the outputs percent and refluxes about the strippers are

done as following:

1-material balance about stripper of heavy naphtha (H.N):

Percent reflux in top tower is constant (6%)

The input amount for strippers towers (H.N) and steam =

𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻. 𝑁 𝑡𝑜𝑡𝑎𝑙

(𝑡𝑜𝑡𝑎𝑙 𝑝𝑒𝑟𝑐𝑒𝑛𝑡% − 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑟𝑒𝑓𝑙𝑢𝑥)× 100

2266560𝐾𝑔𝑑𝑎𝑦

(100 − 6)× 100 = 2411234.043

𝐾𝑔

𝑑𝑎𝑦

The percent of reflux at the top of the column (H.N) and steam= (H.N) Input to

strippers– (input without reflux)

2411234.043𝐾𝑔

𝑑𝑎𝑦− 2266560

𝐾𝑔

𝑑𝑎𝑦= 144674.0426

𝐾𝑔

𝑑𝑎𝑦

𝑇ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑏𝑦 𝑔𝑎𝑙𝑙𝑜𝑛 𝑢𝑛𝑖𝑡 (𝑔𝑎𝑙):

= 144674.0426𝐾𝑔

𝑑𝑎𝑦 ×

1 𝑙𝑖𝑡𝑒𝑟

1 𝐾𝑔 ×

1 𝑔𝑎𝑙

3.785412 𝑙𝑖𝑡𝑒𝑟

= 38218.83658𝑔𝑎𝑙

𝑑𝑎𝑦

The amount steam injected to stripper will be stander for (H.N)

Amount steam = 0.2268 𝐾𝑔

𝑔𝑎𝑙

The amount steam = (H.N) reflux percent ×steam amount = 38218.83658𝑔𝑎𝑙

𝑑𝑎𝑦 ×

0.2268𝐾𝑔

𝑔𝑎𝑙= 8668.032136

𝐾𝑔

𝑑𝑎𝑦

65

Total of reflux Amount from steam and (H.N)

= Amount steam + (H.N) reflux at the top column

=8668.032136𝐾𝑔

𝑑𝑎𝑦+ 144674.0426

𝐾𝑔

𝑑𝑎𝑦= 153342.0747

𝐾𝑔

𝑑𝑎𝑦

The product bottom of (H.N)stripper = input of (H.N)stripper + amount of steam –

reflux amount of steam and (H.N)

= 2411234.043 + 8668.032136 − 153342.0747 = 2266560𝐾𝑔

𝑑𝑎𝑦

Figure 3.3 Material balance Stripper (H.N)

Out put In put

2419902.075 2419902.075

3.6. The Material Balances About The Stripper Of Kerosene (ker):

The percent of reflux at the top of the tower is constant (4%)

The input amount to Kerosene stripper tower= 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑘𝑒𝑟𝑜𝑠𝑒𝑛𝑒

(𝑡𝑜𝑡𝑎𝑙 𝑝𝑒𝑟𝑐𝑒𝑛𝑡%−𝑟𝑒𝑓𝑙𝑢𝑥 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 )× 100

2077680𝐾𝑔𝑑𝑎𝑦

(100 − 4)× 100 = 2164250

𝐾𝑔

𝑑𝑎𝑦

The Percent of reflux at the top of the column (Ker) and steam = Input to strippers

(Ker) – (input without reflux)

66

= 2164250𝐾𝑔

𝑑𝑎𝑦− 2077680

𝐾𝑔

𝑑𝑎𝑦= 86570

𝐾𝑔

𝑑𝑎𝑦

𝑇ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑏𝑦 𝑔𝑎𝑙𝑙𝑜𝑛 𝑢𝑛𝑖𝑡 (𝑔𝑎𝑙) = 86570𝐾𝑔

𝑑𝑎𝑦 ×

1𝑙𝑖𝑡𝑒𝑟

1𝐾𝑔 ×

1𝑔𝑎𝑙

3.785412𝑙𝑖𝑡𝑒𝑟

= 22869.37327𝑔𝑎𝑙

𝑑𝑎𝑦

The amount steam injected to stripper will be stander for kerosene:

𝑎𝑚𝑜𝑢𝑛𝑡 𝑠𝑡𝑒𝑎𝑚 = 0.13608𝐾𝑔𝑠𝑡

𝑔𝑎𝑙

So the amount steam water necessary to strip = Percent reflux (Ker) × amount steam

= 22869.37327𝑔𝑎𝑙

𝑑𝑎𝑦 × 0.13608

𝑘𝑔𝑠𝑡

𝑔𝑎𝑙= 3112.064314

𝐾𝑔 𝑠𝑡𝑒𝑎𝑚

𝑑𝑎𝑦

total reflux Amount from steam and kerosene = amount steam water + reflux kerosene

= 3112.064314𝑘𝑔𝑠𝑡𝑒𝑎𝑚

𝑑𝑎𝑦+ 86570

𝐾𝑔

𝑑𝑎𝑦= 89682.06431

𝑘𝑔

𝑑𝑎𝑦

The product bottom stripper(Kerosene)=

Input stripper (Ker) + column steam water – amount reflux steam and kerosene

= 2164250𝐾𝑔

𝑑𝑎𝑦+ 3112.064314

𝐾𝑔𝑠𝑡𝑒𝑎𝑚

𝑑𝑎𝑦− 89682.06431

𝐾𝑔

𝑑𝑎𝑦= 2077680

𝐾𝑔

𝑑𝑎𝑦

67

Figure3.4 Material balance stripper (kerosene)

Out put In put

2167362.06431 2167362.06431

3.7. Material -Balance about the Stripper of Diesel (diel).

The Percent reflux in top tower is constant (2%)

The input amount to diesel stripper tower (die) and steam

=𝑚𝑎𝑠𝑠 𝑜𝑓 𝑑𝑖𝑒𝑠𝑒𝑙 𝑡𝑜𝑡𝑎𝑙

𝑡𝑜𝑡𝑎𝑙 𝑝𝑒𝑟𝑐𝑒𝑛𝑡% − 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑟𝑒𝑓𝑙𝑢𝑥 × 100

= 4533120

𝑘𝑔𝑑𝑎𝑦

100 − 2 × 100 = 4625632.653

𝑘𝑔

𝑑𝑎𝑦

The Percent of reflux at the top of column (die) and steam =

Input to stripper (diesel) and steam – input without reflux

= 4625632.653𝑘𝑔

𝑑𝑎𝑦− 4533120

𝑘𝑔

𝑑𝑎𝑦= 92512.65306

𝑘𝑔

𝑑𝑎𝑦

68

𝑇ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑏𝑦 𝑔𝑎𝑙𝑙𝑜𝑛 𝑢𝑛𝑖𝑡 (𝑔𝑎𝑙) = 92512.65306𝑘𝑔

𝑑𝑎𝑦 ×

1𝑙𝑖𝑡𝑒𝑟

1𝑘𝑔 ×

1𝑔𝑎𝑙

3.785412𝑙𝑖𝑡𝑒𝑟

= 24439.25603𝑔𝑎𝑙

𝑑𝑎𝑦

The amount steam injected to stripper will be stander for diesel:

𝑎𝑚𝑜𝑢𝑛𝑡 𝑠𝑡𝑒𝑎𝑚 = 0.02265𝑘𝑔

𝑔𝑎𝑙So

amount steam water = percent reflux (diesel) × amount steam

24439.25603𝑔𝑎𝑙

𝑑𝑎𝑦 × 0.02265

𝑘𝑔

𝑔𝑎𝑙= 553.5491492

𝑘𝑔𝑠𝑡𝑒𝑎𝑚

𝑑𝑎𝑦

total reflux Amount from steam and diesel = amount steam water + reflux top

= 553.5491492 + 92512.65306 = 93066.20221𝑘𝑔

𝑑𝑎𝑦

The product bottom column diesel=

Input stripper (diesel) + amount steam water – amount reflux Steam and (diesel)

= 4625632.653 + 553.5491492 − 93066.20221 = 4533120𝑘𝑔

𝑑𝑎𝑦

Figure 3.5 Material balance strippers (diesel)

In put Out put

4626186.2021492 4626186.20221

69

The simulation for these calculations are done by MATLAB available in the appendix

Figure (3.6) Diagram shows the unity of the oil refining and production ratio passes in each

unit

70

CHAPTER FOUR

71

ENERGY BALANCE

When the mass transfer balance derived by applying the law of conservation of mass,

which states that the total generation of mass is 0, also there is one more major conservation

law which provides an additional equation which is the conservation of energy.

4.1. TYPES OF ENERGY`

In order to write an energy balance, we need to know what kinds of energy can

enter or leave a system. Here are some examples (this is not an exhaustive list by any

means) of the types of energy that can be gained or lost.

1. A system could gain or lose kinetic energy, if we're analyzing a moving system.

2. Again, if the system is moving, there could be potential energy changes.

3. Heat could enter the system via conduction, convection, or radiation.

4. Work (either expansion work or shaft work) could be done on, or by, the system.

The total amount of energy entering the system is the sum of all of the different

types entering the system. (8)

4.2. CALCULATION OF ENERGY BALANCE FOR THE HEAT EXCHANGERS

72

The energy balance for all units which use 8 heat exchangers

Table (4-1) initial and final temperature for oil products and their specific heat

Unit Inlet temperature℃ Outlet temperature ℃ specific heat

𝐾𝐽/𝐾𝑔℃

Crude Oil 26.67 - 2.55

Heavy naphtha 141.67 82.22 2.007

Kerosene 180.56 40 2.14

Cold Diesel A ,B 171.11 81.11 2.30

Heavy naphtha

pumproulid A ,B

175.56 97.78 2.007

Cold Residue A,B 254.44 161.11 2.39

Hot Diesel 252.22 182.22 2.30

Diesel pumproulid A

,B

270.56 195.56 2.30

Hot Residue 356.11 254.44 2.39

1-Heavy Naphtha heat exchanger

The temperature of crude oil equal 26.67℃

The temperature of heavy naphtha that inlet to the heat exchanger = 141.67℃

The temperature of heavy naphtha that outlet from the heat exchanger = 82.22℃

And by using the total energy equation Q = mc∆T

Where:

73

m: the mass of product

c: specific heat for product

∆T: temperature difference

(mH.N*cpH.N*∆TH.N) = (moil*cpOil*∆TOil)

(5701476*2.007*(141.67-82.22))= (29733489.6*2.55*∆T)

∆T = 680278165.6 / 75820398.48

∆T=8.97℃

TOil out =TOil int+8.97

TOil out =35.64℃

2- Kerosene heat exchanger

The temperature of kerosene that inlet to the heat exchanger =180.56℃

The temperature of kerosene that outlet from the heat exchanger = 40℃

The temperature of crude oil which inlet to the second heat exchanger = 35.64℃

(mke*cpke*∆T)=(mOil*cpoil*∆T)

(2077680*2.14*(180.56-40))-(29733489.6*2.55*∆T)

∆T= 624962819.7 / 75820398.48

∆T=8.243℃

TOil out =TOil int+8.243

=35.64+8.243

TOil out =43.88℃

3-1 cold diesel (A) heat exchanger

The temperature of cold diesel that inlet to the heat exchanger = 171.11℃

The temperature of cold diesel that outlet from the heat exchanger =81.11℃

The temperature of crude oil inlet to the heat exchanger = 43.88℃

74

(mco Die*cpco Die *∆T) = (mOil *cpOil *∆TOil)

(4533120*2.30*(171.11-81.11)) = (29733489.6*2.55*∆T)

∆T = 938355840 / 75820398.48

∆𝑇 = 12.38℃

TOil out =TOil in +12.38

=43.88 +12.38

TOil out=56.26℃

3-2 cold diesel (B) heat exchanger

QDie =mOil*cpoil*∆𝑇

=29733489.6*2.55*∆T

∆T = 12.38℃

TOil out =Toil in+12.38

=56.26+12.38

TOil out=68.64℃

4-1 Heavy Naphtha pumproulid (A) heat exchanger

The temperature of Heavy Naphtha inlet to the heat exchanger = 175.56℃

The temperature of Heavy Naphtha outlet from the heat exchanger = 97.78℃

The temperature of crude oil inlet to the heat exchanger = 68.64℃

mH.N*cpH.N*∆𝑇= mOil*cpOil*∆𝑇

5701476*2.007*(175.56-97.78)=29733489.6*2.55*∆𝑇

∆T = 890025832.2 / 75820398.48

∆𝑇 = 11.74

TOil out= Toil in+11.74

TOil out =80.38℃

75

4-2 Heavy Naphtha pumproulid heat (B) exchanger

QH.N=mOil *cpOil*∆𝑇

890025832.2=29733489.6*2.55*∆𝑇

∆𝑇 = 11.74℃

TOil out =80.38+11.74

=92.12℃

5-1 the cold residue (A) heat exchanger

The temperature of cold residue inlet to the heat exchanger = 254.44℃

The temperature of cold residue outlet from the heat exchanger = 161.11℃

The temperature of crude oil inlet to the heat exchanger = 92.12℃

mC.Re*cpC.Re*∆𝑇=mOil *cpOil *∆𝑇

22612713.6*2.39*(254.44-161.11) =29733489.6*2.55*∆𝑇

∆T = 5043962499 / 75820398.48

∆𝑇 = 66.53℃

Toil out =92.12+66.53

=158.65℃

5-2 the cold residue (B) heat exchanger

QC.Re= mOil *cpOil *∆𝑇

5043962499=29733489.6*2.55*∆𝑇

∆𝑇 = 66.53℃

Toil out=158.65+66.53

Toil out=225.18℃

6- Hot Diesel heat exchanger

The temperature of hot diesel inlet to the heat exchanger = 252.22℃

76

The temperature of hot diesel outlet from the heat exchanger = 182.22℃

The temperature of the crude oil inlet to the heat exchanger = 225.18℃

mH.De*cpH.De*∆𝑇=mOil*cpOil*∆𝑇

4533120*2.30*(252.22-182.22 ) =29733489.6*2.55*∆𝑇

∆T = 729832320 / 75820398.48

∆𝑇 = 9.63℃

TOil out =225.18+9.63

TOil out=234.81℃

7-1 Diesel pumparoulid (A) heat exchanger

The temperature of diesel pumparoulid inlet to the heat exchanger = 270.56℃

The temperature of diesel pumparoulid outlet from the heat exchanger = 195.56℃

The temperature of diesel pumparoulid inlet the to heat exchanger = 234.81℃

mDe.P*cpDe.P*∆𝑇=mOil*cpOil*∆𝑇

4533120*2.30*(270.56-195.56)=29733489.6*2.55*∆𝑇

∆T = 781963200 / 75820398.48

∆𝑇 = 10.3℃

TOil out=234.81+10.3

TOil out=245.12℃

7-2 Diesel pumparoulid (B) heat exchanger

QDe.p=mOil*cpOil*∆𝑇

781963200=29733489*2.55*∆𝑇

∆𝑇 = 10.3℃

TOil out=245.12+10.3

TOil out=255.4℃

77

8- Hot Residue heat exchanger

The temperature of hot residue inlet to the heat exchanger = 356.11℃

The temperature of hot residue outlet from the heat exchanger = 54.44℃

The temperature of crude oil inlet to the heat exchanger = 255.4℃

mH.De*cpH.De*∆𝑇=mOil*cpOil*∆𝑇

22612713.6*2.39*(356.11-254.44) = 29733489.6*2.55*∆𝑇

∆T = 5494692674 / 75820398.48

∆𝑇 = 72.47℃

TOil out =255.4+72.47

TOil out =327.79℃ this is the final temperature of crude oil

Figure(4.1)shows increase temperature crude oil inside heat exchanger

Line Product

Oil

Heavy naphtha

Kerosene

78

Diesel

Radium

Table (4-2) temperature crude oil in each unit

Heat Exchanger temperature of out oil (KJ)

E1

E2

E3A

E3B

E4A

E4B

E5A

E5B

E6

E7A

E7B

E8

35.64

43.88

56.26

68.64

80.38

92.12

158.65

225.18

234.81

245.12

255.4

327.79

Table (4-3) the overall energy balance

Qin Qout

680278165.6

624962819.7

938355840

890025832.2

5043962499

72983232

78196320

5494692674

680108974.4

624760083.5

938656533.2

890131478.2

5044331111

729771335.4

781935769.5

5494628457

1.382*1010

1.382*1010

Q in = Q out

1.382*1010

= 1.382*1010

Qin Q out

Crude unit

79

80

CHAPTER FIVE

DESIGN OF SHELL&TUBE HEAT EXCHANGER

A shell and tube heat exchanger is a class of heat exchanger designs.[1][2]

It is

the most common type of heat exchanger in oil refineries and other large chemical

processes, and is suited for higher-pressure applications. As its name implies, this

type of heat exchanger consists of a shell (a large pressure vessel) with a bundle of

tubes inside it. One fluid runs through the tubes, and another fluid flows over the

tubes (through the shell) to transfer heat between the two fluids. The set of tubes is

called a tube bundle, and may be composed of several types of tubes: plain,

longitudinally finned, etc. (9)

5.1. SHELL AND TUBE HEAT EXCHANGER DESIGN

There can be many variations on the shell and tube design. Typically, the ends

of each tube are connected to plenums (sometimes called water boxes) through holes

in tubesheets. The tubes may be straight or bent in the shape of a U, called U-tubes.

Fig (5-1) U-tube heat exchanger

In nuclear power plants called pressurized water reactors, large heat exchangers

called steam generators are two-phase, shell-and-tube heat exchangers which typically

81

have U-tubes. They are used to boil water recycled from a surface condenser into

steam to drive a turbine to produce power. Most shell-and-tube heat exchangers are

either 1, 2, or 4 pass designs on the tube side. This refers to the number of times the

fluid in the tubes passes through the fluid in the shell. In a single pass heat exchanger,

the fluid goes in one end of each tube and out the other. (10)

Fig (5-2) Straight tube heat exchanger (one pass tube)

Surface condensers in power plants are often 1-pass straight-tube heat exchangers

(see surface condenser for diagram). Two and four pass designs are common because

the fluid can enter and exit on the same side. This makes construction much simpler.

Fig (5-3) Straight tube heat exchanger (two pass tube)

There are often baffles directing flow through the shell side so the fluid does

not take a short cut through the shell side leaving ineffective low flow volumes. These

82

are generally attached to the tube bundle rather than the shell in order that the bundle

is still removable for maintenance.

Counter current heat exchangers are most efficient because they allow the highest log

mean temperature difference between the hot and cold streams. Many companies

however do not use two pass heat exchangers with a u-tube because they can break

easily in addition to being more expensive to build. Often multiple heat exchangers

can be used to simulate the counter current flow of a single large exchanger. (9)

5.1.1. Selection of tube material

To be able to transfer heat well, the tube material should have good thermal

conductivity. Because heat is transferred from a hot to a cold side through the tubes,

there is a temperature difference through the width of the tubes. Because of the

tendency of the tube material to thermally expand differently at various

temperatures, thermal stresses occur during operation. This is in addition to

any stress from high pressures from the fluids themselves. The tube material also

should be compatible with both the shell and tube side fluids for long periods under

the operating conditions (temperatures, pressures, pH, etc.) to minimize deterioration

such as corrosion. All of these requirements call for careful selection of strong,

thermally-conductive, corrosion-resistant, high quality tube materials,

typically metals, including aluminum, copper alloy, stainless steel, carbon steel, non-

ferrous copper alloy, Inconel, nickel, Hastelloy and titanium] Fluoropolymerssuch

as Perfluoroalkoxy alkane (PFA) and Fluorinated ethylene propylene (FEP) are also

used to produce the tubing material due to their high resistance to extreme

temperatures.[4]

Poor choice of tube material could result in a leak through a tube

between the shell and tube sides causing fluid cross-contamination and possibly loss

of pressure. (9)

5.1.2. Applications and uses

The simple design of a shell and tube heat exchanger makes it an ideal cooling

solution for a wide variety of applications. One of the most common applications is

the cooling of hydraulic fluid and oil in engines, transmissions and hydraulic power

packs. With the right choice of materials they can also be used to cool or heat other

83

mediums, such as swimming pool water or charge air. One of the big advantages of

using a shell and tube heat exchanger is that they are often easy to service, particularly

with models where a floating tube bundle (where the tube plates are not welded to the

outer shell) is available. (9)

5.2. SHELL AND TUBE EXCHANGERS: CONSTRUCTION

DETAILS

The principal types of shell and tube exchanger are shown in Figures to

The standard nomenclature used for shell and tube exchanger is given below:

1. Shell 15. Floating-head support

2. Shell cover 16. Weir

3. Floating-head cover 17. Split ring

4. Floating-tube plate 18. Tube

5. Clamp ring 19. Tube bundle

6. Fixed-tube sheet (tube plate) 20. Pass partition

7. Channel (end-box or header) 21. Floating-head gland (packed gland)

8. Channel cover 22. Floating-head gland ring

9. Branch (nozzle) 23. Vent connection

10. Tie rod and spacer 24. Drain connection

11. Cross baffle or tube-support plate 25. Test connection

12. Impingement baffles 26. Expansion bellows

13. Longitudinal baffle 27. Lifting ring

14. Support bracket

Fig (5-4) U-tube heat exchanger constructions

84

5.3. STEPS FOR DESIGN OF HEAT EXCHANGER:

Design a shell-and-tube exchanger for the following data:

2077680kg/day =86570kg/h of kerosene (42°API) leaves the base of a kerosene side-

stripping column at 180.56℃ and is to be cooled to 40℃ by exchange with

37723113.6kg/day=1571796.4kg/h light crude oil (34°API) coming from storage at

33.74℃. The kerosene enters the exchanger at a pressure of 5 bar and the crude oil at

6.5 bar. A pressure drop of 0.8 bar is permissible on both streams. Allowance should

be made for fouling by including a fouling factor of 0.0003 (𝑊/𝑚2℃)−1on the crude

stream and 0.0002 (𝑊/𝑚2℃)−1on the kerosene stream.

An algorithm for the design of shell-and tube exchangers is shown in Figure 4.2 the

procedure set out in this figure will be followed in the design.

Fig(5-5) Design procedures for shell-and-tube heat exchangers

85

Step 1: Specification

The specification is given as following:

86570kg/h of kerosene (42° API) at 180.56℃ cooled to 40℃, by exchange with

1571796.4kg/h light crude oil (34° API) at 33.74℃.

The kerosene pressure 5 bar, the crude oil pressure 6.5 bar.

Permissible pressure drops of 0.8 bars on both streams.

Fouling factors: crude stream 0.00035(𝑊/𝑚2℃)−1,

Kerosene stream 0.0002(𝑊/𝑚2℃)−1

To complete the specification, the duty (heat transfer rate) and the outlet temperature

of the crude oil needed to be calculated.

The mean temperature of the kerosene =180.56+40

2= 110.28℃

At this temperature the specific heat capacity of 42◦ API kerosene is 2.14 kJ/kg℃

𝑄𝑘𝑒𝑟 =86570

3600× 2.14(180.56 − 40) = 7233.365969𝑘𝑊

As a first trial take the mean temperature of the crude oil as equal to the inlet

temperature, 33.74℃; specific heat capacity at this temperature = 2.55 kJ/kg℃.

An energy balance gives:

1571796.4

3600× 2.55(𝑡2 − 33.74) = 1509.4

𝑡2 = 35.09℃so the stream mean temperature = (33.74 + 35.09)/2 = 34.415℃

The specific heat at this temperature is 2.05 kJ/kg℃

86

Step 2: Physical Properties:

Tab (5-1) shows Physical Properties for oil and kerosene.

Kerosene Inlet Mean Outlet

Temperature 180.56 110.28 40 ℃

specific heat 2.72 2.47 2.26 kJ/kg℃

Thermal conductivity 0.130 0.132 0.135 W/m℃

Density 690 730 770 kg/m3

Viscosity 0.22 0.43 0.80 mN sm−2

Crude oil Outlet Mean inlet

Temperature 37.179 35.45 33.74 ℃

specific heat 2.09 2.05 2.01 kJ/kg℃

Thermal conductivity 0.133 0.134 0.135 W/m℃

Density 800 820 840 kg/m3

Viscosity 2.4 3.2 4.3 mN sm−2

87

Step 3: Overall coefficient

For an exchange of this type the overall coefficient will be in the range

300 to 500𝑊/𝑚2 ℃, between heavy oils and steams; so start with 300

𝑊/𝑚2 ℃

Step 4: Exchanger type and dimensions

An even number of tube passes is usually the preferred arrangement, as these

positions the inlet and outlet nozzles at the same end of the exchanger, which

simplifies the pipework.

Start with one shell pass and 2 tube passes.

∆𝑇𝑙𝑚 =(180.56 − 35.09) − (40 − 33.74)

ln(180.56 − 35.09)

(40 − 33.74)

= 44.253℃

𝑅 =(𝑇1 − 𝑇2)

(𝑡2 − 𝑡1)

𝑅 =(180.56 − 40)

(35.09 − 33.74)= 104.11

𝑆 =(35.09 − 33.74)

(180.56 − 33.74)= 0.0096

88

Fig (5-6) temperature correction factors

From Figure 4.3 temperature correction factor one shell pass two or more even tube

passes𝐹𝑡 = 0.75, which is acceptable.

So, ∆𝑇𝑚 = 0.75 × 44.253 = 33.18975℃

Step 5: Heat transfer area

𝐴𝑜 =1509.4 × 103

300 × 33.18975= 151.593 𝑚2

Step 6: Layout and tube size

Using a split-ring floating head exchanger for efficiency and ease of cleaning

Neither fluid is corrosive, noris the operating pressure not high, so plain carbon steel

can be used for the shell and tubes.

The crude is dirtier than the kerosene, so put the crude through the tubes and the

kerosene in the shell.

Use 19.05 mm (3/4 inch) outside diameter, 14.83 mm inside diameter, 5 m Long tubes

on a triangular 23.81 mm pitch.

89

Step 7: Number of tubes

Area of one tube (neglecting thickness of tube sheets)

= 𝜋 × 19.05 × 10−3 × 5 = 0.2992 𝑚2

Number of tubes = A heat transfer / A one tube =151.593 /0.2992 = 506.66

So, for 2 passes, tubes per pass = 325 tubes

Check the tube-side velocity at this stage to see if it looks reasonable.

Tube cross-sectional area =𝜋

4(14.83 × 10−3)2 = 0.0001727 𝑚2

So area per pass = 325 × 0.0001727 = 0.05612 𝑚2

Volumetric flow =1571796.4

3600×

1

820= 0.5325 𝑚3/𝑠

Tube-side velocity, ut =Qoil

A tube one pass =

0.5325

0.05612= 9.4877 m/s

The velocity is satisfactory, between 3 to 5m/s, but may be a little low. This will show

up when the pressure drop is calculated.

Step 8: Bundle and shell diameter

Table (5-2) Constant for use in equation tube arrangement

90

Equivalent diameter, cross-sectional areas and wetted perimeters

From Table 5.2, for 2 tube passes,K1 = 0.249, n1 = 2.207

So, 𝐷𝑏 = 19.05 (506.66

0.249)

1

2.207= 601.17 𝑚𝑚 (0.601 𝑚)

Fig (5-7) Shell-bundle clearances

91

For a split-ring floating head exchanger the typical shell clearance from Figure (4-4)

is 63 mm, so the shell inside diameter,

𝐷𝑠 = 601.17 + 63 = 664.17𝑚𝑚

Step 9: Tube-side heat transfer coefficient

𝑅𝑒 =

= 820×9.4877×14.83×10−3

3.2×10−3 = 0.03605 (0.03605 /1000=3605)

𝑃𝑟 = = 2.05 × 103 × 3.2 × 10−3

0.134= 48.96

𝐿

𝑑𝑖=

5000

14.83= 337

From Figure (4-5), so reading is 𝑗ℎ = 3.2 × 10−3

Fig (5.8) Tube-side heat-transfer factor

92

𝑁𝑢 = 3.2 × 10−3(0.03605)(48.96)0.33 = 0.0004166(0.4166)

ℎ𝑖 = 0.4166 × (0.134

14.83 × 10−3) = 3.76(376) 𝑊/𝑚2 ℃

This is clearly too low if U◦ is to be 300 W/m2℃. The tube-side velocity did look

low, so increase the number of tube passes to 4. This will halve the cross-sectional

area in each pass and double the velocity.

New

𝑢𝑡 = 2 × 9.4877 = 18.9754 𝑚/𝑠

And

𝑅𝑒 = 2 × 0.03605 = 0.0721

𝑗ℎ = 3.2 × 10−3

ℎ𝑖 = (0.134

14.83 × 10−3) × 3.8 × 10−3(0.0721)(48.96)0.33 = 8939 𝑊/𝑚2 ℃

Step 10: Shell-side heat transfer coefficient

Kern’s method will be used.

With 4 tube passes the shell diameter will be larger than that calculated for 2 passes.

For 4 passes K1 =0.175 and n1 =2.285

𝐷𝑏 = 19.05 (506.66

0.175)

1

2.285= 623.5𝑚𝑚 (0.623 𝑚)

The bundle to shell clearance is still around 56 mm, giving:

𝐷𝑠 = 623.5 + 56 = 679.5 𝑚𝑚

93

As a first trial take the baffle spacing = 𝐷𝑠/5, say 150 mm. This spacing should give

good heat transfer without too high a pressure drop.

𝐴𝑠 =(23.81 − 19.05)

23.81× 679 × 135.8 = 18433.89 𝑚𝑚2 = 0.01843𝑚2

𝑑𝑒 =1.10

19.05(23.812 − 0.917 × 19.052) = 13.52 𝑚𝑚

Volumetric flow-rate on shell-side =86570

3600×

1

730= 0.0329

𝑚3

𝑠

Shell-side velocity = q/𝐴𝑠 =0.0329

0.01843= 1.785 𝑚/𝑠

𝑅𝑒 =730 × 1.620 × 13.52 × 10−3

0.43 × 10−3= 37183.14(3.718 × 104)

𝑃𝑟 =2.14 × 103 × 0.43 × 10−3

0.132= 6.97

Use segmental baffles with a 25% cut. This should give a reasonable heat transfer

coefficient without too large a pressure drop.

From Figure(4-5) , 𝑗ℎ = 4.52 × 10−3.

Neglecting the viscosity correction:

ℎ𝑠 = (0.132

13.52× 103) × 4.52 × 10−3 × 37183.14 × 6.970.33 = 3114 𝑊/𝑚2 ℃

94

Step 11: Overall coefficient

The overall coefficient calculation from this relation:

1

𝑈𝑜= (

1

8939+ 0.00035)

19.05

14.83+

19.05 × 10−3 ln (19.0514.83)

2 × 55+

1

3266+ 0.0002

𝑈𝑜 = 863.93 𝑊/𝑚2 ℃

This is above the initial estimate of 300 𝑊/𝑚2 ℃. The number of tubes could

possibly be reduced, but first check the pressure drops.

Step 12: Pressure drop Tube-side

651 tubes, 4 passes, tube inside diameter 14.83 mm,𝑢𝑡 = 9.487 m/s,

𝑅𝑒 = 7.2 × 103. From Figure (5.9),so reading 𝑗𝑓 = 5 × 10−3.

Fig (5-9) Tube-side friction factors

95

∆𝑃𝑡 = 4 (8 × 5 × 10−3 (5000

14.83) + 2.5)

(820 × 9.4872)

2

= 4(13.5 + 2.5)(820 × 902)

2

= 23596 𝑁/𝑚2 ,2.3 𝑏𝑎𝑟

5.4. MODIFIED DESIGN

The tube velocity needs to be reduced. This will reduce the heat transfer

coefficient, so the number of tubes must be increased to compensate. There will be a

pressure drop across the inlet and outlet nozzles. Allow 0.1 bars for this, a typical

figure (about 15% of the total); which leaves 0.7 bars across the tubes. Pressure drop

is roughly proportional to the square of the velocity and 𝑢𝑡 is proportional to the

number of tubes per pass. So the pressure drop calculated for 240 tubes can be used to

estimate the number of tubes required.

Tubes needed = 240/(0.6/1.4)0.5 = 366

Say, 360 with 4 passes.

Retain 4 passes as the heat transfer coefficient will be too low with 2 passes.

Second trial design: 360 tubes 19.05 mm out diameter, 14.83 mm inside diameter, 5 m

long, triangular pitch 23.81 mm.

Db = 19.05 (360

0.175)

12.207

= 604 mm (0.604 m)

96

From Figure (4-4) clearance with this bundle diameter D 59 mm

𝐷𝑠 = 604 + 59 = 663 𝑚𝑚

Cross-sectional area per pass =360

4(14.83 × 10−3)2 𝜋

4= 0.01555 𝑚2

Tube velocity 𝑢𝑡 =𝑞

𝐴=

0.0237

0.01555= 1.524

𝑚

𝑠

𝑅𝑒 =820 × 1.524 × 14.83 × 10−3

3.2 × 10−3= 5792

L/d is the same as the first trial, 337

𝑗ℎ = 3.6 × 10−3

ℎ𝑖 = (0.134

14.83× 103) × 3.6 × 10−3(5792)(48.96)0.33 = 680 𝑊/𝑚2 ℃

This looks satisfactory, but checks the pressure drop before doing the shell-side

calculation.

𝑗𝑓 = 5.5 × 10−3

∆𝑃𝑡 = 4 (8 × 5.5 × 10−3 (5000

14.83) + 2.5)

(820 × 1.5242)

2

= 66,029 𝑁/𝑚2 , 0.66 𝑏𝑎𝑟

Well within specification.

Keep the same baffle cut and spacing.

𝐴𝑠 =(23.81 − 19.05)

23.81× 663 × 100 = 13254 𝑚𝑚2 = 0.01325 𝑚2

𝑢𝑠 =0.0076

0.01325 = 0.574 𝑚/𝑠

𝑑𝑒 = 13.52 𝑚𝑚, 𝑎𝑠 𝑏𝑒𝑓𝑜𝑟𝑒

𝑅𝑒 =730 × 0.574 × 13.52 × 10−3

0.43 × 10−3= 13175 , (1.3 × 104)

𝑃𝑟 = 8.05

97

𝑗ℎ = 4.8 × 10−3 , 𝑗𝑓 = 4.6 × 10−3

ℎ𝑠 = (0.132

13.52 × 10−3) × 4.8 × 10−3 × 13175 × (6.97)0.33 = 1172 𝑊/𝑚2 ℃

∆𝑃𝑠 = 8 × 4.6 × 10−3 (663

13.52) (

5000

14.83)

(730 × 0.5742)

2

= 73169.5 𝑁/𝑚2 , 1.2 𝑏𝑎𝑟

Too high; the specification only allowed 0.8 overall, including the loss over the

nozzles.

Check the overall coefficient to see if there is room to modify the shell-side design.

1

𝑈𝑜= (

1

683+ 0.00035)

19.05

14.83+

19.05 × 10−3 ln (19.0514.88)

2 × 55+

1

1172+ 0.0002

𝑈𝑜 = 291.79 𝑊/𝑚2 ℃

𝑈𝑜 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 =𝑄

(𝐴𝑜∆𝑇𝑙𝑚), 𝐴𝑜 = 360 × 0.2992 = 107.7 𝑚2,

𝑠𝑜 𝑈𝑜 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 =1509.4 × 103

(107.7 × 44.253)= 316 𝑊/𝑚2 ℃

The estimated overall coefficient is well above that required for design, 302 compared

to 192𝑊/𝑚2 ℃, which gives scope for reducing the shell-side pressure drop.

Allow a drop of 0.1 bars for the shell inlet and outlet nozzles, leaving 0.7 bar for the

shell-side flow. So, to keep within the specification, the shell-side velocity will have

to be reduced by around√(1/2) = 0.707. To achieve this baffle spacing will need to

be increased to 100/0.707 = 141, say 140 mm.

As =(23.81 − 19.05)

23.81× 663 × 140 = 1855 mm2 = 0.186 m2

us =0.0076

0.186 = 0.408 m/s

98

Giving: 𝑅𝑒 = 10,443, ℎ𝑠 = 1177 𝑊/𝑚2 ℃, ∆𝑃𝑠 = 0.47 𝑏𝑎𝑟,and 𝑈𝑜 =

288 𝑊/𝑚2 ℃.

The pressure drop is now well within the specification.

Step 13: Estimate cost

The cost of this design can be estimated on the value of parts to construct it.

Step 14: Optimization

There is scope for optimizing the design by reducing the number of tubes, as the

pressure drops are well within specification and the overall coefficient is well above

that needed.

However, the method used for estimating the coefficient and pressure drop on the

shell-side (Kern’s method) is not accurate, so keeping to this design will give some

margin of safety.

Summary: the proposed design

Split ring, floating head, 1 shell pass, and 4 tube passes.

360 carbon steel tubes, 5 m long, 19.05 mm out diameter, 14.83 mm inside diameter,

triangular pitch,

Pitch 23.18 mm.

Heat transfer area 107.7 𝑚2 (based on outside diameter).

Shell inside diameter 597 mm (600 mm), baffle spacing 140 mm, 25% cut.

Tube-side coefficient 680𝑊/𝑚2 ℃, clean

Shell-side coefficient 1366𝑊/𝑚2 ℃,, clean.

Overall coefficient, estimated 288𝑊/𝑚2 ℃, dirty.

Overall coefficient required 197𝑊/𝑚2 ℃, dirty.

Dirt/Fouling factors:

Tube-side (crude oil) 0.00035 (𝑊/𝑚2 ℃)−1.

Shell-side (kerosene) 0.0002 (𝑊/𝑚2 ℃)−1.

Pressure drops:

99

Tube-side, estimated 0.40 bar, +0.1 for nozzles; specified 0.8 bar overall.

Shell-side, estimated 0.45 bar, +0.1 for nozzles; specified 0.8 bar overall.

5.5. OPTIMIZATION USING A CAD PROGRAM

The use of a proprietary computer program (HTFS, M-TASC) to find the

lowest cost design that meets the specification resulted in the design set out below.

The program selected longer tubes, to minimize the cost. This has resulted in an

exchanger with a shell length to diameter ratio of greater than 10: 1. This could cause

problems in supporting the shell, and in withdrawing the tube bundle for maintenance.

The CAD program was rerun with the tube length restricted to 3500 mm, to produce a

more compact design. This gave a design with 349 tubes, 4 passes, in a shell 540 mm

diameter. The setting plan for this design is shown in Figure.

100

Fig (5-10) Setting out plan for compact design. (Courtesy of Heat Transfer and Fluid

Flow Service, Harwell)

Split ring, floating head, 1 shell pass, 2 tube passes.

168 carbon steel tubes, 6096 𝑚𝑚, 19.05 𝑚𝑚 o.d., 14.83 𝑚𝑚 i.d., triangular pitch,

pitch 23.18 𝑚𝑚.

Heat transfer area 61 𝑚2.

Shelli.d 387, baffle spacing 77.9𝑚𝑚, 15% cut.

Tube-side coefficient 851 𝑊/𝑚2 ℃, clean.

Shell-side coefficient 1191 𝑊/𝑚2 ℃, clean.

Overall coefficient estimated 484 (𝑊/𝑚2 ℃)−1 clean.

Overall coefficient estimated 368 (𝑊/𝑚2 ℃)−1dirty.

Pressure drops, including drop over nozzles:

Tube-side, estimated 0.5 𝑏𝑎𝑟.

Shell-side, estimated 0.5 𝑏𝑎𝑟.

5.6 SHAPE PARTS OF THE HEAT EXCHANGER

101

Fig (5-11) the tube arrangement

Fig (5-10) The tube geometry

102

Fig (5-11) Types of baffle used in shell and tube heat exchangers. (a) Segmental (b)

Segmental and strip (c) Disc and doughnut (d) Orifice

103

Fig (5-12) Clearance and flow areas in the shell-side of a shell and tube exchanger

104

105

CHAPTER SIX

ECONOMIC STUDY

6.1. FEASIBILITY STUDY

A feasibility study is an analysis of how successfully a project can be completed,

accounting for factors that affect it such as economic, technological, legal and scheduling

factors. Project managers use feasibility studies to determine potential positive and negative

outcomes of a project before investing a considerable amount of time and money into it. A

feasibility study aims to objectively and rationally uncover the strengths and weaknesses of

an existing business or proposed venture, opportunities and threats present in

the environment, the resources required to carry through, and ultimately the prospects for

success. In its simplest terms, the two criteria to judge feasibility are cost required

and value to be attained. A well-designed feasibility study should provide a historical

background of the business or project, a description of the product or service, accounting

statements, details of the operations and management, marketing research and policies,

financial data, legal requirements and tax obligations. Generally, feasibility studies precede

technical development and project implementation. A feasibility study evaluates the

project's potential for success; therefore, perceived objectivity is an important factor in the

credibility of the study for potential investors and lending institutions. It must therefore be

conducted with an objective, unbiased approach to provide information upon which

decisions can be based. A feasibility study tests the viability of an idea, a project or even a

new business. The goal of a feasibility study is to place emphasis on potential problems that

could occur if a project is pursued and determine if, after all significant factors are

considered, the project should be pursued. Feasibility studies also allow a business to

address where and how it will operate, potential obstacles, competition and the funding

needed to get the business up and running.(11)

106

6.2. PLANT COST

It is often necessary to estimate the cost of a piece of plant when no cost data are

available for the particular size of operational capacity involved. Good results can be

obtained by using the logarithmic relationship known as the six-tenths-factor rule, if the new

piece of plant is similar to one of another capacity for which cost data are available.

According to this rule, if the cost of a given unit at one capacity is known, the cost of a

similar unit with X times the capacity of the first is approximately (X) m times the cost of the

initial unit.(12)

m

Cost of plant B = cost of plant A (capac.of plant B capac.of plant A

)

m= 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑦𝑒𝑎𝑟−𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑦𝑒𝑎𝑟

100

In this project there a comparison the prices of the sarrir refinery that have capacity

of 20.000 barrel per day with a plant have capacity 200.000 barrel per day, which recover

the order of gasoline stations in tobruck city, The cost of plant B must be converting from

the original year (1989) to present year (2018) depending on the cost index to obtain the

fixed cost. And the cost of plant A= 1.3*108$, m= 0.27

0.27

Cost of plant B = 1.3*108 (200.000

20.000 ) = 2.4* 10 8 $

6.3. COST INDEX

Most cost data which are available for immediate use in a preliminary or predesign

estimate are based on conditions at some time in the past. Because prices may change

considerably with time due to changes in economic conditions, some method must be used

for updating cost data applicable at a past date to costs that are representative of conditions

107

at a later time; this can be done by the use of cost indexes. A cost index is merely an index

value for a given point in time showing the cost at that time relative to a certain base time. If

the cost at some time in the past is known, the equivalent cost at the present time can be

determined by multiplying the original cost by the ratio of the present index value to the

index value applicable when the original cost was obtained.Cost indexes can be used to give

a general estimate, but no index can take into account all factors, such as special

technological advancements or local conditions. The common indexes permit fairly accurate

estimates if the time period involved is less than 10 years.

present cost = original cost ∗(index value at present time)

(index value at time original cost was obtained)

Cost index at present time (2018) = (1090)

Cost index at original time (1989) = (800)

Present cost (2016) = 2.4*10 8 *

(1090)

(800)

Fixed cost = Present cost = 3.27*10 8 $

Assume the variable cost equal 60% of fixed cost

Variable cost = 0.6 * fixed cost = 1.9*10 8

$

Total cost = fixed cost +Variable cost = 5.03*10 8$

108

6.4. THE COST OF PRODUCTS AND RAW MATERIAL

The next table shows the price of the products and the flow rate for each product by

cubic meter per day, data taken from mass balance and the price for the year 2018, assume

the products prices equal 50$/m3

Table (6-2) the flow rate of products and their prices.(13)

Products

Price $/m3

m3/Day

LPG 50 528.864

Light naphtha 50 5704.176

Heavy naphtha 50 2266.56

Kerosene 50 2077.68

Diesel 50 4533.12

Residue 50

22612.7136

The cost of the products = (the flow rate for the product1* its price + product2 *its price ….)

Assume the working house for the units is 7200 hr/year, which equal 300 day/year

The cost of the products/ day = [(528.864*50) + (5704.176*50) + (2266.56*50) +

(2077.68*50) + (4533.12*50) + (22612.7136*50)]*300

= 5.65*108 $/year

The next table shows the flow rate for raw materials and their

Cost, assume the price of the raw material equal 10$/m3

109

Table (6-2) the flow rate of the raw material and their prices.(13)

Raw material

$/m3

Flow rate m3/day

Crude oil

10

37776

The cost of the RWM = (the flow rate for the RWM 1* its price + RWM 2 *its price...)

The cost of RWM /year = [(37776*10)] * 300 = 1.13*108 $/year

6.5. DEPRECIATION

Depreciation is an accounting method of allocating the cost of a tangible asset over

its useful life. Businesses depreciate long-term assets for both tax and accounting purposes.

For tax purposes, businesses can deduct the cost of the tangible assets they purchase as

business expenses; however, businesses must depreciate these assets in accordance with

IRS rules about how and when the deduction may be taken.

Assume 30 years depreciation period and the installation has no salvage value

Depreciation =total cost (C)

life period (n)

Depreciation =5.03 ∗ 108

30= 1.7 ∗ 107 $/𝑦𝑒𝑎𝑟

110

6.6. THE COST OF THE OPERATION

The cost of operation is the business strategy implemented in many companies to gain

a huge market. Cost of operation is the cost acquired in completing one operation. It may be a

conversion of inputs into the outputs or labor costs etc. If the cost of operation is low then it is

easy to maintain cost leadership and gain the market with competitive advantage.

Operating cost = Raw material (RM) + utilities (util) + operation labor + others

RM + DEP = 90% from production cost

Production cost =RM + DEP

0.9

Production cost =1.13 ∗ 108 + 1.7 ∗ 107

0.9= 1.44 ∗ 108 $/𝑦𝑒𝑎𝑟

Util + Labor = 0.1 * production cost

Util + Labor = 0.1 *1.44 ∗ 108 = 1.44 ∗ 107 $/year

Operating cost =1.44 ∗ 108 + 1.44 ∗ 107 = 1.59*108 $/year

6.7. CASH FLOW

Cash flow is the net amount of cash and cash-equivalents moving into and out of a

business. Positive cash flow indicates that a company's liquid assets are increasing, enabling it

to settle debts, reinvest in its business, return money to shareholders, pay expenses and

provide a buffer against future financial challenges. Negative cash flow indicates that a

company's liquid assets are decreasing. Net cash flow is distinguished from net income, which

includes accounts receivable and other items for which payment has not actually been

received. Cash flow is used to assess the quality of a company's income, that is, how liquid it

is, which can indicate whether the company is positioned to remain solvent.

111

Cash Flow Analysis

Cash flows are often transformed into measures that give information e.g. on a

company's value and situation:

To determine a project's rate of return or value. The time of cash flows into and out of

projects are used as inputs in financial models such as internal rate of return and net

present value.

To determine problems with a business's liquidity. Being profitable does not

necessarily mean being liquid. A company can fail because of a shortage of cash even

while profitable.

As an alternative measure of a business's profits when it is believed that accrual

accounting concepts do not represent economic realities. For instance, a company

may be notionally profitable but generating little operational cash (as may be the case

for a company that barters its products rather than selling for cash). In such a case, the

company may be deriving additional operating cash by issuing shares or raising

additional debt finance.

Cash flow can be used to evaluate the 'quality' of income generated by accrual

accounting. When net income is composed of large non-cash items it is considered

low quality.

To evaluate the risks within a financial product, e.g., matching cash requirements,

evaluating default risk, re-investment requirements, etc.

Cash flow = Net profit

Years 1-30 Prices $/year

Income 5.65*108

Operating -1.59*108

Depreciation -1.7*107

The net profit =3.88*108

Cash flow = 3.88*108$/year

6.8. PAY- BACK PERIOD

112

Payback period is the time in which the initial cash outflow of an investment is

expected to be recovered from the cash inflows generated by the investment. It is one of the

simplest investment appraisal techniques.

Formula

The formula to calculate payback period of a project depends on whether the cash

flow per period from the project is even or uneven. In case they are even, the formula to

calculate payback period is:

pay − back period =Total cost

cash flow

pay − back period =5.03 ∗ 108

3.88 ∗ 108= 1.297 𝑦𝑒𝑎𝑟

Advantages of Payback Period are:

1. Payback period is very simple to calculate.

2. It can be a measure of risk inherent in a project. Since cash flows that occur later in a

project's life are considered more uncertain, payback period provides an indication of how

certain the project cash inflows are.

3. For companies facing liquidity problems, it provides a good ranking of projects that

would return money early.

Disadvantages of Payback Period are:

1. Payback period does not take into account the time value of money which is a serious

drawback since it can lead to wrong decisions. A variation of payback method that

attempts to remove this drawback is called discounted method.

2. It does not take into account, the cash flows that occur after the payback period. (11)

113

114

CONCLUSION

1. It is crucial to take into account the importance of having oil refinery producing units

in Libya.

2. A maximizing factor to the oil refinery producing unit of Sarir must be taken into

account to have one in Brega city.

3. The classification of the crude oil is an important factor for the unit design

4. The material balance for this unit was 200,000 barrel of oil per day

5. The overall energy balance for the heat exchangers 1, 2…..8 was 1.382*1010KJ.

6. The cost of Brega refinery was ……..

RECOMMINDATIONS

1. The designing of the heater is recommended for further study.

2. Applying the same study for different crude oils.

3. Adding a chapter for the process control.

115

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Petroleum Refineries,” Jan. 5, 1994

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(3)-Peach leaf, Jaeger Products, Inc., Houston, Texas, http://www.jaeger.com,2016

(4)-http://www.ogj.com/seb/2016

(5)-Brix Jacobsen -Department Head, Structural and Fluid Mechanics, R&T- book: The

Centrifugal Pump, RESEARCH AND TECHNOLOGY, www.grundfos.com, 2006

(6)-Paresh Girdhar B. Eng. (Mech Eng.), Practical Centrifugal Pumps -Design, Operation and

Maintenance, www.integra-india.com, 2005

(7)-Himmelblau, David M. Basic Principles and Calculations in Chemical Engineering (2nd

ed.). Prentice Hall. (1967).

(8)-Europa http://ec.europa.eu/eurostat/web/energy/data/energy-balances, 2016

Haifa, Oil Refineries Ltd, Israel, GALEC@ORL.CO.IL,2016

(9)-Surinder Park ash, Refining Processes Handbook, United States of America,

http://www.gulfpp.com, 2003

(10)-Bytocom , http://www.bytocom.com/vb/t31198.html,2015

(11)-www.investopedia.com/22/10/2016

(12)-Klaus D. Timmerhaus, Professors of Chemical Engineering University of Colorado,

Plant Design And Economics For Chemical Engineers, 1991

(13)-Sarir field, Libya, 2016

(14)-Albadwe, www.unep.org/pcfv/PDF/AlbadweYahaConf.pdf- , 2015

(15)-Aris. A normalization for the Thiele modulus. Ind. Eng. Chem. Fundam., 4:227, 1965.

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ISBN 978-1-934768-59-4. (1971).

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of Illinois at Chicago , USA, http:/ /www.springer.com, 2006

116

(18)-James G. Speight, CD&W Inc., The Chemistry and Technology of Petroleum,

Wyoming, http://www.taylorandfrancis.com, 2007

(19)-James H. Gary, Colorado School of Mines & Glenn E. Handwork, Consulting Chemical

Engineer, Petroleum Refining Technology and Economic, http:/ /www.dekker.com, 2001

(20)-John jechura, jechura@ mines.edu , gasoline upgrading, 2015

(21)-Khalid Abdullah, Dr. in Omear Al-Mukhtar university, corrosion control, 2015

(22)-Mahmoud Ibrahim, Arabian heath institute, Damascus, 1988

\(23)-Raheel G. Raheel, Dr., in Tobruk university ,Petroleum Refinery, 2016

(24)-Reyad Shawabkeh, Dr. Department of Chemical Engineering, King Fahd University of

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(25)-Riazi, M.R. Characterization and Properties of Petroleum Fractions, ASTM, 2005.

(26)-Richard L. Myers, The 100 most important chemical compounds /a reference guide/

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117

Appendix A

Simulation to mass balance Unit by MATLAB

1- % material balance

2- function [x]= matblan(oilvol,sgoil)

3- clc

4- format short

5- syms oilmas

6- oilmas= solve(oilvol*159==oilmas);

7- dw=1000;

8- doil=sgoil*dw;

9- API=(141.5/sgoil)-131.5;

10- voloil=oilmas/doil;

11- pg1=4.25*0.33;

12- ln1=6;

13- hn1=15.1;

14- ker1=25*0.22;

15- die1=38.5*0.31;

16- res1=41*1.46;

17- %the mass of product

18- vlpg=voloil*(pg1/100);

19- vln=voloil*(ln1/100);

20- vhn=voloil*(hn1/100);

21- vker=voloil*(ker1/100);

22- vdie=voloil*(die1/100);

23- vres=voloil*(res1/100);

24- vtotal=(vlpg+vln+vhn+vker+vdie+vres);

25- dlpg=650; dln=740; dhn=780;

26- dker=790; ddie=840; dres=900;

27- mlpg=vlpg*dlpg;

118

28- mln=vln*dln;

29- mhn=vhn*dhn;

30- mker=vker*dker;

31- mdie=vdie*ddie;

32- mres=vres*dres;

33- %material bulence on strippres

34- unit=(1/3.785412);

35- sghn=0.78; sgker=0.79; sgdes=0.84;

36- sglpg=0.65; sgres=0.90; sgln= 0.74;

37- mlpg1=mlpg/sglpg;

38- mln1=mln/sgln;

39- mhn1=mhn/sghn;

40- mker1=mker/sgker;

41- mdie1= mdie/sgdes;

42- mres1=mres/sgres;

43- mtotal=mlpg1+mln1+mhn1+mdie1+mker1+mres1;

44- %heavy naphata

45- hnstrp=(mhn1/(100-6))*100;

46- refluxhntop=hnstrp-mhn1;

47- refluxhntop1=refluxhntop*unit;

48- stem=0.2268;

49- amountstemh2o=refluxhntop1*stem;

50- totalrefluxhn=amountstemh2o + refluxhntop;

51- trefluxtop=totalrefluxhn + refluxhntop;

52- %theproduct botem strippres H.N

53- prodstrihn=((hnstrp + amountstemh2o) - trefluxtop);

54- %steripper Kerosen

55- kerstrp=(mker1/(100-4))*100;

56- refluxtop=kerstrp - mker1;

57- refluxker1=refluxtop*unit;

58- amountstem=0.13608;

59- stemh2o=refluxker1*amountstem;

119

60- treflux=stemh2o + refluxtop;

61- %proudect of kerosen

62- proker=(kerstrp + stemh2o) - treflux;

63- %desele

64- dieinp=(mdie1*(100/98));

65- diereflx=dieinp-mdie1;

66- diereflx1=diereflx*unit;

67- amountstemdie=0.02265;

68- stmh2o=amountstemdie*diereflx1;

69- totalref=stmh2o + diereflx;

70- %prudect

71- prdcolum=(dieinp + stmh2o) - totalref;

72- disp('the API=')

73- double(API)

74- disp('mass of oil=')

75- double(mtotal)

76- disp('mass of LPG=')

77- double(mlpg1)

78- disp('mass of Light naphtha =')

79- double(mln1)

80- disp('mass of Heavy naphtha =')

81- double(mhn1)

82- disp('mass of Kerosene =')

83- double(mker1)

84- disp('mass of Diesel =')

85- double( mdie1 )

86- disp('mass of Residue =')

87- double(mres1)

88- disp('the mass of heavy naphtha inlet the striper=')

89- double(hnstrp)

90- disp('the total reflux H.N=')

91- double(trefluxtop)

120

92- disp('the product from strip H.N=')

93- double(prodstrihn)

94- disp('the mass of Kerosene inlet the striper=')

95- double(kerstrp)

96- disp('the total reflux Ker=')

97- double(treflux)

98- disp('the product from strip Ker=')

99- double(proker)

100- disp('the mass of Diesel inlet the striper=')

101- double(dieinp)

102- disp('the total reflux Diesel=')

103- double(totalref)

104- disp('the product from strip Die=')

105- double(prdcolum)

121

Appendix B

Simulation To The Energy Balance By MATLAB

1- % energy balance

2- function [x]=

3- enblan(to1,mo,mhn,mk,md,mre)

4- clc

5- thn1=141.67;

6- thn2=82.22;

7- cpo=2.55;

8- cphn= 2.007;

9- dthn=thn1-thn2;

10- syms dto

11- dto = solve ((mhn*cphn*dthn)==(mo*cpo*dto));

12- syms to2

13- to2=solve (dto==to2-to1);

14- cpk=2.14;

15- tk1=180.56;

16- tk2=40;

17- dtk=(tk1-tk2);

18- syms dto1

19- dto1= solve((mk*cpk*dtk)==(mo*cpo*dto1));

20- syms to3

21- to3 =solve (dto1==to3-to2);

22- cpd=2.30;

23- td1=171.11; td2=81.11;

24- dtd=(td1-td2);

25- syms dto2

26- dto2= solve ((md*cpd*dtd)==(mo*cpo*dto2));

122

27- syms to4

28- to4= solve(dto2==to4-to3);

29- syms dtod

30- dtod= solve((md*cpd*dtd)==(mo*cpo*dtod));

31- syms to5

32- to5= solve(dtod==to5-to4);

33- thnp1=175.56; thnp2=97.78;

34- dthnp=(thnp1-thnp2);

35- syms dto3

36- dto3= solve((mhn*cphn*dthnp)==(mo*cpo*dto3));

37- syms to6

38- to6= solve(dto3==to6-to5);

39- syms dto3b

40- dto3b= solve ((mhn*cphn*dthnp)==(mo*cpo*dto3b));

41- syms to6b

42- to6b= solve (dto3b==(to6b-to6));

43- tcr1=254.44; tcr2=161.11;

44- cpre=2.39;

45- dtcr=(tcr1-tcr2);

46- syms dt4

47- dt4=solve ((mre*cpre*dtcr)==(mo*cpo*dt4));

48- syms to7

49- to7=solve(dt4==to7-to6b);

50- syms dt4b

51- dt4b= solve((mre*cpre*dtcr)==(mo*cpo*dt4b));

52- syms to7b

53- to7b=solve (dt4b==to7b-to7);

54- thd1=252.22; thd2=182.22;

55- dthd=(thd1-thd2);

56- syms dt5

57- dt5= solve((md*cpd*dthd)==(mo*cpo*dt5));

58- syms to8

123

59- to8= solve(dt5==to8-to7b);

60- tdp1=270.56; tdp2=195.56;

61- dtdp= tdp1-tdp2;

62- syms dto6

63- dto6= solve((md*cpd*dtdp)==(mo*cpo*dto6));

64- syms to9

65- to9= solve(dto6==to9-to8);

66- syms dto6b

67- dto6b= solve((md*cpd*dtdp==(mo*cpo*dto6b)));

68- syms to9b

69- to9b= solve(dto6b==to9b-to9);

70- thr1=356.11; thr2=254.44;

71- dthr=thr1-thr2;

72- syms dto7

73- dto7= solve((mre*cpre*dthr)==(mo*cpo*dto7));

74- syms to10

75- to10= solve(dto7==to10-to9b);

toi=[to1,to2,to3,to4,to5,to6,to6b,to7,to7b,to8,to9,to9b,to10]

76- numhx=[1,2,3,4,5,6,6.5,7,7.5,8,9,9.5,10]

77- plot(toi, numhx)

78- grid

79- title('the temperature of oil')

80- %toi lable ('Number of H.Ex')

81- format short

82- disp('the temp for oil inlet EX1')

83- double(to1)

84- disp('the temp for oil inlet EX2')

85- double(to2)

86- disp('the temp for oil inlet EX3')

87- double(to3)

88- disp('the temp for oil inlet EX3b')

89- double(to4)

124

90- disp('the temp for oil inlet EX4')

91- double(to5)

92- disp('the temp for oil inlet EX4b')

93- double(to6)

94- disp('the temp for oil inlet EX5')

95- double(to6b)

96- disp('the temp for oil inlet EX5b')

97- double(to7)

98- disp('the temp for oil inlet EX6')

99- double(to7b)

100- disp('the temp for oil inlet EX7')

101- double(to8)

102- disp('the temp for oil inlet EX7b')

103- double(to9)

104- disp('the temp for oil inlet EX8')

105- double(to9b)

106- disp('the temp for oil out the unit')

107- double(to10)