CPE604 - Mini Project Plant Design (Production Of Phenol) (2018)

PLANT DESIGN FOR PRODUCTION OF PHENOL

FACULTY OF CHEMICAL ENGINEERING UNIVERSITI TEKNOLOGI MARA TITLE: MINI DESIGN PROJECT (PRODUCTION OF PHENOL)

PREPARED BY: EH2205I 1

MUHAMMAD NOR ASYRAFF BIN NOR RAHMAN

2017632104

2

PUTERA NAJMEEN FARITH BIN ABDUL RAZAK

2017632096

3

NURUL NAJIHA BINTI SURANI

2017632166

4

NURUL AMIRAH BINTI MUSDAFA KAMAL

2017632124

5

SITI NUR AISHAH BINTI MOHAMAD FAUDZI

2017632136

6

NURUL AIDA BINTI MOHAMMAD

2017632132

7

NURLINA SYAHIIRAH BINTI MD TAHIR

2017632214

8

QURRATUAINI BINTI MD ALI

2017632078

9

NURSYAWANI YASMIN BINTI RAMLI

2017632094

10

JULIA BINTI FADZLI

2017632064

DATE OF SUBMISSION: 14th DECEMBER 2018 NAME OF LECTURER: EN AMMAR BIN MOHD AKHIR

i


TABLE OF CONTENTS HISTORY AND GENERAL INFORMATION ON PHENOL ............................................ 1

1.0 1.1

History.................................................................................................................................... 1

1.1.1

Phenol ............................................................................................................................. 1

1.1.2

The Production Of Phenol Synthesis ........................................................................... 2

1.2

General Description .............................................................................................................. 5

1.3

Usage ...................................................................................................................................... 8 PROCESS SELECTION ........................................................................................................ 10

2.0 2.1

Process in Producing Phenol .............................................................................................. 10

2.1.1 2.2

Process Selected and Selection Criteria .................................................................... 10

Profit Margin ....................................................................................................................... 16

2.2.1

Phenols from Toluene Benzoic Acid .......................................................................... 16

2.2.2

Phenols From Coal ...................................................................................................... 17

2.2.3

Phenol From Cumene Peroxidation .......................................................................... 18

3.0

SITE SELECTION ................................................................................................................. 20

3.1

Kerteh, Terengganu ................................................................................................................ 23 3.1.1

Market Availability ..................................................................................................... 23

3.1.2

Transportation ............................................................................................................ 23

3.1.3

Availability Of Labor.................................................................................................. 27

3.1.4

Utilities and Facilities.................................................................................................. 28

3.1.5

Land Estimation .......................................................................................................... 29

3.1.6

Climate ......................................................................................................................... 30

3.1.7

Waste Management..................................................................................................... 30

3.2

Gebeng, Pahang................................................................................................................... 32

3.2.1


3.2.2

Transportation ............................................................................................................ 33

3.2.3


3.2.4


3.2.5

Land Estimation .......................................................................................................... 34

3.2.6

Climate ......................................................................................................................... 36

3.2.7


3.3

Pasir Gudang Industrial Estate, Johor Bahru ................................................................. 37

3.3.1


3.3.2

Transportation ............................................................................................................ 37

3.3.3


ii

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.3.4


3.3.5

Land Estimation .......................................................................................................... 39

3.3.6

Climate ......................................................................................................................... 40

3.3.7


4.0

BLOCK FLOW DIAGRAM FOR PHENOL PLANT ........................................................ 41

5.0

PROCESS FLOW DIAGRAM FOR PHENOL PLANT .................................................... 42

6.0

MATERIALS BALANCE ...................................................................................................... 44

6.1

Reactor 1 – Oxidation Reactor .......................................................................................... 44

6.2

Reactor 2 - Cleavage Unit................................................................................................... 46

6.3

Separator 1 - Phase Separator (Gravity Settler) .............................................................. 48

6.4

Separator 2 – Scrubber....................................................................................................... 49

6.5

Distillation Unit ................................................................................................................... 50

6.5.1

Separator 3 – Distillation Unit 1 ................................................................................ 51

6.5.2


6.5.3


6.5.4


6.5.5


6.6

Reactor 3 – Pre - Purification Reactor .............................................................................. 56 ENERGY BALANCE ............................................................................................................. 57

7.0 7.1

Sample Calculation ............................................................................................................. 57

7.1.1

Pump 1 – Cumene Feed Pump ................................................................................... 58

7.1.2

Compressor 1 – Air Feed ............................................................................................ 59

7.1.3

Reactor 1 – Oxidation Reactor .................................................................................. 60

7.1.4

Reactor 2 – Cleavage Unit .......................................................................................... 62

7.1.5

Heat Exchanger 1 – Cooling ....................................................................................... 63

7.1.6

Separator 1 - Phase Separator (Gravity Settler) ...................................................... 64

7.1.7

Separator 2 – Scrubber............................................................................................... 65

7.1.8


7.1.9


7.1.10


7.1.11


7.1.12


7.1.13


7.1.14


8.0 8.1

SIZING OF EQUIPMENTS .................................................................................................. 74 Sizing Of The Equipments ................................................................................................. 74

iii

PLANT DESIGN FOR PRODUCTION OF PHENOL 8.1.1

Reactor 1 – Oxidation Reactor CSTR ....................................................................... 75

8.1.2

Reactor 2 – Cleavage Unit CSTR .............................................................................. 76

8.1.3

Reactor 3 – Pre-Purification Reactor CSTR ............................................................ 77

8.1.4

Separator 1 - Phase Separator (Gravity Settler) ...................................................... 77

8.1.5

Separator 2 - Scrubber ............................................................................................... 80

8.1.6

Separator 3 – Distillation Column Unit 1 ................................................................. 82

8.1.7


8.1.8


8.1.9


8.1.10


9.0

COSTING OF EQUIPMENTS ............................................................................................ 104

9.1

Reactor 1 – Oxidation Reactor CSTR ............................................................................. 104

9.2

Reactor 2 – Cleavage Unit CSTR .................................................................................... 104

9.3

Reactor 3 – Pre-Purification Reactor CSTR .................................................................. 104

9.4

Separator 1 – Phase Separator (Gravity Settler) ........................................................... 105

9.5

Separator 2 – Scrubber..................................................................................................... 107

9.6

Separator 3 – Distillation Column Unit 1 ....................................................................... 109

9.7


9.8


9.9


9.10

Separator 7 – Distillation Column Unit 5 ...................................................................... 122

9.11

Total Equipment Cost....................................................................................................... 126

10.0

OPERATING LABOUR COST .......................................................................................... 127

12.0

CONCLUSION ..................................................................................................................... 129

BIBLIOGRAPHY ............................................................................................................................. 130

iv


0


1.0

HISTORY AND GENERAL INFORMATION ON PHENOL

1.1

History

1.1.1 Phenol Phenol is a constituent of coal tar and was partly isolated from coal tar in 1834 by Runge, who called it ‘Carbolic Acid’ or ‘coal oil acid’. Friedlieb Ferdinand Runge who is born on 8 February 1795, began his career in Breslau, Germany. Runge published several scientific and technological papers and books (Megson.N.J.L 1958). Pure Phenol was first prepared by Laurent in 1841. Auguste Laurent born 4 September 1808, the son of a wine merchant, was assistant to Dumas at the Ecole Centrale (1831) and to Brongniart at the Sevres porcelain factory (1833–1835) in France. Since 1845, he worked in a laboratory at the Ecole Normale in Paris. In his studies of the distillate from coal-tar and chlorine, Laurent isolated Dichlorophenol, C24H8Cl4O2 and trichlorophenol (acide chlorophénisique) C24H6Cl6O2, which both suggested the existence of phenol. He give it the name phène. He provided the table of ‘general formulae’ of the derived radicals of phène where phenol was indicated by the incorrect formula C24H12 + H4O2 (C6H8O, in modern notation) (L.Ross 2003). In 1841, Laurent isolated and crystallized phenol for the first time. He called it ‘hydrate de phenyle’ or ‘acide phenique’. His reported the Melting point between 34 and 35˚C and Boiling point of 187 and 188 ˚C which are rather similar with the values we known today. Lauren also measure the elementary physical properties of the chemical. Lauren try to apply the phenyle as a painkiller for the toothache. The effect on the pain was rather unclear, but the substance was ‘very aggressive on the lips and the gums’. In the analysis of his experiments, Laurent applied the substitution hypothesis that was originally proposed by his former supervisor, Dumas. However, Laurent went further than Dumas and assumed that the substitution reaction did not change the structural formula of the reactant and the product, whereas Dumas limited himself to the claim that the removal of one hydrogen atom was replaced by the addition of another group, leaving open the possibility of a complete rearrangement of the molecule (F.L Tobiason 2012). Yet, the substitution hypothesis (especially in the form proposed by Laurent) was attacked rather strongly by Berzélius, who claimed that a simple replacement of the hydrogen

1

PLANT DESIGN FOR PRODUCTION OF PHENOL atom should be utterly impossible. Instead, he reinterpreted all the results of Laurent by breaking up the reaction product into smaller molecules. In 1843, Charles Frederic Gerhardt also prepared phenol by heating salicylic acid with lime and gave it the name ‘phenol’ (Looney 1995). He subsequently published a paper on ‘The Negative Nature Of The Phenyl Group’, where he noted how phenyl together with other ‘negative groups’ can make the hydrogen atoms in methylene groups more reactive. In 1867, Heinrich von Brunck defended his Ph.D. thesis on the theme ‘About Derivatives of Phenol’, where he particularly studied the isomers of nitrophenol. The Raschig–Dow process of manufacturing phenol by cumene was discovered by Wurtz and Kekule in 1867, although the earlier synthesis was recorded by Hunt in 1849. Interestingly, Friedrich Raschig, working earlier as a chemist at BASF and known for his work on the synthesis of phenol, later established his own company in Ludwigshafen (L.Ross 2003). During World War I, Phenol was misuse by the Nazis in 1942 for execution by injection into the prisoners. They justify their action as an experimental procedure towards human. 1.1.2 The Production Of Phenol Synthesis Phenol can be produced from many sources such as cumene, toluene and benzene with different processes. There are several processes technology associated to Phenol production, which are:i)

Cumene Peroxidation Process

Benzene and purified propylene obtained from petroleum industry are mixed in liquid or vapour phase in presence of phosphoric acid on kieselguhr. As a result, cumene or isopropyl benzene is formed. The cumene thus formed is made into the form of an emulsion with dilute aqueous sodium carbonate solution, using sodium stearate as an emulsifier. The emulsion is then oxidized in an oxidizer with air under atmospheric pressure for 3-4 hours in presence of a catalyst, such as copper, cobalt or manganese salt. The temperature and pH of the reaction are maintained between 160-260°C and 8.5-10.5, respectively. As a result of oxidation, cumene hydroperoxide is formed. The peroxide thus formed is then decomposed by 5-50% sulphuric acid in an acidifier at 45-65°C under pressure. As a result of decomposition, phenol (15%), acetone (9%), cumene (73%) are formed along with some α-methylstyrene and acetophenone. These separated by a separator. The cumene is recycled to be used again and phenol is either extracted or recovered by distillation. The yield is about 92%. Acetone is formed as a by-product (0.6 lb. per lb of phenol).

2

PLANT DESIGN FOR PRODUCTION OF PHENOL ii) This

Rasching Process process

was

developed

in

Germany

in

1940.

Benzene

is

first converted into chlorobenzene by passing a mixture of benzene vapour, hydrochloric acid vapour and air under normal pressure at about 23°C in presence of a copper iron catalyst, supported on alumina. The reaction is exothermic in nature and so the temperature is maintained constant by external cooling. The conversion .the per pass is 10%. The chlorobenzene after separation from unchanged reactants is hydrolyzed into phenol by heating with steam at about 400-500°C in presence of silica catalyst. The conversion is again about 10% per pass in this second step. Hydrogen chloride set free in the reaction is recovered and recycled. Crude phenol (97%) obtained according to the above reaction is purified by distillation under vacuum. The yield is about 75-85% of benzene. A small amount of HCl is sufficient to convert large amounts of benzene into phenol.

iii)

Toluene Two-Stage Oxidation Process

Toluene in the liquid phase is oxidized with air in a reactor under 40-70psi in presence of a soluble cobalt catalyst maintained at 150°C. Benzoic acid and water are thus formed. The reaction is exothermic and the temperature is maintained by external cooling. The crude molten benzoic acid at about 150-200°C is transferred from the reactor to distillation column, where separation of benzoic acid from unreacted toluene and produced water take place. The toluene is separated and recycled to the first oxidizing reactor. The pure benzoic acid is fed to a second reactor, where it is oxidized to phenol by air and steam under 20-25psi at 230°C in presence of cupric benzoate catalyst promoted with manganese. The reaction mass is periodically withdrawn from the second reactor into an extractor, where it is washed with water to remove unwanted tars and benzoic acid and steam are returned to the second reactor. The phenol, water and unreacted benzoic acid are conducted overhead to two distillation columns in series. In the first column, crude phenol is separated from overhead and unreacted benzoic acid is recycled to the second oxidizing reactor. Pure form phenol is obtained at the second distillation column as overhead product and supply's aromatics compounds and benzoic acid as a feed for crude phenol rectification column. The yield of phenol on benzoic acid is about 75- 80%.

3

PLANT DESIGN FOR PRODUCTION OF PHENOL iv)

Sulphonation Process

It is one of the oldest methods of manufacture of phenol. Benzene sulphonic acid is first prepared by passing vapour of benzene into concentrated sulphuric acid is about 150-170°C. The water formed during sulphonation process is distilled out because sulfuric acid gets diluted and conditions accelerate backward reaction to the process. Benzene sulphonic acid should be neutralized by reacting it with aqueous sodium sulphite to form the salt of benzene sulphonic acid. The sodium salt is filtered off and then fused with caustic in a cast iron vessel at about 340-380°C in the ratio (1:3) for about 5-6 hours. As a result, sodium phenate is formed. The melt is cooled, extracted with water and then acidified with sulphur dioxide. The latter is obtained as a result of neutralization of benzene sulphonic acid with sodium sulphite. The upper oily layer of crude phenol is distilled under vacuum to get pure phenol. The yield is about 8090% of benzene. The lower layer contains sodium sulphite which is separated and used for the neutralization of benzene sulphonic acid. Table 1 Shows the Differences between Processes to Produce Phenol No

Existing Methods

1.

Cumene Peroxidation

2.

Toluene TwoStage Oxidation

3.

Rasching Phenol Process

4.

Chlorobenzene caustic hydrolysis

5.

Benzene sulfonate process

Raw Materials / Catalyst Cumene, Air, Sulfuric acid and Emulsifying agents Toluene, Air, Cobalt napthalate, Cupric benzoate catalyst Benzene, Air, Hydrochloric acid

Yield

Products

Comments

92%

Phenol and acetone

80%

Phenol and Carbon dioxide

75%

Benzene, Chlorine, Sodium hydroxide, Hydrochloric acid Benzene, Sulfuric acid , Sodium hydroxide

95%

Phenol and Hydrochloric acid as recycling Phenol, Sodium chloride Phenol, Sodium sulphite, Sodium sulphate

Produces valuable coproduct acetone Low-cost operation by direct toluene application Feasible under large units

87%

Economically not feasible Operates on large batch cycle

4


1.2

General Description

A phenol is an organic compound in which an -OH group is attached to a carbon atom that is part of an aromatic carbon ring system. The name phenol is derived from a combination of the term phenyl and alcohol.

Figure 1 Shows Structure Of Phenols The general formula for phenols is Ar-OH, where Ar represents an aryl group. An aryl group is an aromatic carbon ring system from which one hydrogen atom has been removed. The reaction chemistry for phenols differs from non-aromatic alcohols to justify these compounds separately. Phenols contain a “benzene ring” and the chemistry of benzene differs from other unsaturated hydrocarbons. The followings are examples of compound that classified as phenols:

Figure 2 Shows Classification Of Phenols For the naming of phenols simply depends on the IUPAC rules where the extensions of the rules used to name benzene derivatives with hydrocarbon or halogen substituents. The parent name is phenol. Ring numbering always begins with the hydroxyl group and proceeds in the direction which gives lower number next to the carbon atom bearing a substituent. The numerical position of the hydroxyl group is not specified in the name because it is 1 by definition (H. Stephen Stoker; 2016).

5


Figure 3 shows examples of names of phenols given by the IUPAC rules

Figure 4 shows methyl and hydroxy derivatives of phenols, cresol

Figure 5 shows hydroxyphenols three isomers with different names

Physical and Chemical Properties of Phenols Phenols are generally low-melting solids or oily liquids at room temperature. Most phenols are slightly soluble in water. Many phenols have antiseptic and disinfectant properties. The simplest phenol, phenol itself, is a colorless solid with a medicinal odor. Its melting point is 41°C, and is more soluble in water than are most other phenols.

6

PLANT DESIGN FOR PRODUCTION OF PHENOL Basically, the chemical properties of phenols differ from those of alcohols. The similarities and differences between these two reactions are as follows: Table 2 Shows Similarities And Differences Between Alcohols And Phenols Alcohols

Relation

Phenols

Flammable

Similarities

Flammable

Undergoes halogenation Dehydrates

Undergoes halogenation Differences

Undergoes oxidation

Does not dehydrates Does not undergoes oxidation

Acidity of Phenols One of the important properties of phenols is their acidity. Unlike alcohols, phenols are weak acids in solution. As acids, phenols have Ka values of about 10-10. Such Ka values are lower than those of most weak inorganic acids (10-5 to 10-10). The acid ionization reaction for phenol itself is:

Figure 6 shows ionization reaction for phenol itself

Meanwhile, for negative ion produced from ionization is called phenoxide ion. When phenol itself is reacted with sodium hydroxide (a base), salt sodium phenoxide is produced.

Figure 7 Shows Process For Phenoxide Ion

7


1.3

Usage

There are a few uses of phenol in terms of health, industrial and daily uses. Firstly, the uses of Bisphenol A (BPA). Bisphenol A was created from a condensation reaction of phenol and acetone with hydrogen chloride, an acid catalyst and a promoter such as methyl mercaptan. Once formed by this reaction, BPA is washed with water, neutralized with calcium hydroxide and distilled under vacuum. BPA can be purified further by distillation and extractive crystallization. Higher purity BPA is used to make polycarbonate plastic while the lower purity BPA is used to make epoxy resin. Many water bottles are polycarbonate plastic made of BPA (Kennepohl, 2017). Besides that, BPA is especially used for the production of high grade polycarbonates for compact disc, glazing and for automotive industry (Elyers, 2011). The second largest consumption of phenol is for the production of phenolic resins with formaldehyde. There are mainly used for underseal applications in the automotive industry (Elyers, 2011). The phenolic resins may be considered to be the first polymeric products produced commercially from simple compounds of low molecular weight i.e they were the first truly synthetic resins to be exploited. Furthermore, phenolic resins continue to be used for a wide variety of applications such as moulding powders, laminating resins, adhesives, binders, surface coatings and impregnates. Until very recently the market has continued to grow but not at the same rate as for plastic materials in general. For example, in 1957 production of phenolic resins was of the same order as for PVC and polyethylene and about twice that of polystyrene. Today it is less than a tenth that of polyethylene and about one-third that of polystyrene (Brydson, 1999). Phenols also are widely used as antiseptics (substances that kill microorganisms on living tissue) and as disinfectants (substance intended to kill microorganisms on inanimate objects such as furniture or floors). The first widely used antiseptic was phenol however phenol is toxic to humans which can cause severe burns when applied to the skin. One of the safer phenolic antiseptic is 4-hexylresorcinol (4-hexyl-1,3-dihydroxybenzene; resorcinol is the common name for 1,3-dihydroxybenzene, and 4-hexylresorcinol). It is much more powerful than phenol as a germicide and has fewer undesirable side effects. Indeed it is safe enough to be used as the active ingredients in some mouthwashes and throat lozenges (Kennepohl, 2017). As the automotive industry is an important consumer of phenol derivatives such as polycarbonates and phenolic resins, the entire phenol demand still tends to be cyclical and closely tied to the market. The dominant synthetic route is cumene oxidation, which accounts

8

PLANT DESIGN FOR PRODUCTION OF PHENOL for about 95% of the total phenol production. This process will maintain its leading position as long as a market exists for the coproduct acetone. The classic synthetic routes such as sulfonation or chlorination of benzene are no longer because of the formation of considerable quantities of sodium salts as byproducts. Commercial production of phenol by direct oxidation of benzene with byproduct nitrous oxide does not seem to be viable in the near future (Elyers, 2011). Table 3 Worldwide production capacity and demand for phenol (in 1000 t/a). 2000

2003

2007

2010

Production capacity from Cumene

7641

8490

9879

10079

Toluene

307

307

307

307

Coal tar

291

291

291

291

Total

8239

9088

10477

10677

Demand for Bisphenol A

2306

2678

3446

3864

Phenolic resins

2236

2419

2564

2709

Caprolactam

739

705

724

728

Others

1335

1454

1576

1686

Total

6616

7256

8306

8987

9


2.0

PROCESS SELECTION

2.1

Process in Producing Phenol

2.1.1 Process Selected and Selection Criteria This section will explain the reasons why the selected process has been chosen and the disadvantages of the unselected processes.

In this world there are various ways of producing phenol. Obviously, all the processes have their own advantages and disadvantages. This considerations is important to choose the most efficient and profitable process in order to have a good and nearly perfect production of phenol. Here are the lists of the processes in producing phenol.

a) Production of phenol from cumene b) Benzene Sulfonation c) Oxidation of Toluene d) Phenol from coal e) Toluene Benzoic Acid Process

For this mini project, we have decided to choose production of phenol from cumene. This is the route from cumene to produce phenol. 𝐂𝟔 𝐇𝟓 𝐂𝐇(𝐂𝐇𝟑 ) + 𝐎𝟐 → 𝐂𝟔 𝐇𝟓 𝐂(𝐂𝐇𝟑 )𝟐 𝐎𝐎𝐇 Cumene + Oxygen Gas → Cumeneperoxide 𝐂𝟔 𝐇𝟓 𝐂(𝐂𝐇𝟑 )𝟐 𝐎𝐎𝐇 + 𝐇𝟐 𝐒𝐎𝟒 ↔ 𝐂𝟔 𝐇𝟓 𝐎𝐇 + 𝐂𝐇𝟑 𝐂𝐎𝐂𝐇𝟑 Cumeneperoxide + Sulfuric Acid → Phenol + Acetone Phenol can also be produced by free radical chain reaction of cumeneperoxide. The side product from this reaction is Acetone which also high demand in the industry. Cumeneperoxide can be obtained from air oxidation of cumene (isopropyl benzene). Approximately over 90% cumene produced in industry, in order to satisfies the feedstock for phenol production. An article entitled ‘Industrial Catalytic Processes – Phenol Production’ written by Robert J.

10

PLANT DESIGN FOR PRODUCTION OF PHENOL Shcmidt explain the details of the reaction and the benefits associates with the process (Schmidt, 2005).

The Sunoco/UOP Phenol process produces high-purity phenol and acetone by the cumene peroxidation route, using oxygen from air. This process features low-pressure oxidation for improved yield and safety, advanced CHP cleavage for high product selectivity, an innovative direct product neutralization process that minimizes product waste, and an improved, low cost product recovery scheme. The process also produces an ultra-high product quality at relatively low capital and operating costs. The main reactions for phenol and acetone production via cumene peroxidation are both reactions are highly exothermic.

Oxidation of cumene to cumene hydroperoxide (CHP) proceeds via a free-radical mechanism that is essentially auto-catalyzed by CHP. The decomposition reaction is catalyzed by strong mineral acid and is highly selective to phenol and acetone. The unstable peroxide and many side reactions which take place simultaneously with the above reactions are controlled and minimized by optimization of the process conditions. Dimethylphenylcarbinol is the main oxidation by-product, and the DMPC/AMS reactions play a significant role in the plant. The figure 8 below shows block flow diagram in using the Sunoco/UOP.

Figure 8 Sunoco/UOP Phenol process.

11

PLANT DESIGN FOR PRODUCTION OF PHENOL The major processing steps include: 1. liquid-phase oxidation of cumene to cumene hydroperoxide (CHP) 2. concentration of CHP 3. acid-catalyzed decomposition of concentrated CHP to phenol and acetone 4. neutralization of acidic decomposition product 5. fractionation of the neutralized decomposition product for recovery of acetone, phenol, AMS, and residue 6. recovery of phenol and the effluent wastewater via an extraction process to prepare it for further downstream treatment required to meet effluent quality specifications 7. Hydrogenation of AMS back to cumene for recycling to synthesis or refining of AMS for sale as a product.

Here are the disadvantages of the other processes that have not been selected. 2.1.1.1 Benzene Sulphonation The main disadvantage was the poor atom economy (36.7%) of the reaction (which is the percent ratio of molecular mass of the desired product to the molecular mass of reactant). This means that considerably less than half the mass of reactants ends up in the required product, even if we assume 100% yield. In practice the yield is more likely to be in the region of 88%, giving an atom economy of 32.3%, barely a third of the reactant mass. Sodium sulphite is consumed in the wood pulp and paper industries but the large amount of waste produced was one of the reasons why the benzene sulphonation route is no longer used (Sayyar, 2008).

Figure 9 - The mechanism of the benzene sulfonation process.

12

PLANT DESIGN FOR PRODUCTION OF PHENOL 2.1.1.2 Oxidation Of Toluene There are two types reaction in chlorination of benzene: 1. Alkaline Hydrolysis. According to the (Ullmann, 2005) , the chlorobenzene being formed from the chlorinated of benzene at 38°C to 608°C in the presence of ferric chloride catalyst in the process. The chlorobenzene hydrolyzed with caustic soda (NaOH) at temperature of 4008°C and 2.56 kPa to form sodium phenoxide and sodium cloride. The impure sodium phenoxide reacts with hydrochloric acid to release the phenol from the sodium salt. The yield of phenol is about 82% to that of the theoretical value based on benzene. Typically, the process combined with chloroalkali electrolysis to recover chlorine from the sodium chlorid (Kirk-Othmer, 2006).

Figure 10 The Alkaline Hydrolysis Of Benzene In Production Of Phenol

2. Raschig-Hooker Process. The Raschig-Hooker Process is oxychlorination of benzene reaction which benzene is oxychlorinated with hydrochloric acid, air, and with the presence of iron and copper chloride catalyst to form chlorobenzene. Based on the (Kirk-Othmer, 2006) .The reaction occurs at 200–2608°C and atmospheric pressure. The chlorobenzene hydrolyzed at 4808°C in the presence of a suitable catalyst to produce phenol and hydrochloric acid. Thus, the hydrochloric acid is recycled and the process called as regenerative Raschig process. The yield of phenol is 90 mol% of theoretical (Ullmann, 2005)

Figure 11 The Raschig-Hooker Process In Production Of Phenol

13

PLANT DESIGN FOR PRODUCTION OF PHENOL 2.1.1.3 Phenol From Coal For this process, there are many disadvantages that lead to the inefficient production. Obviously, this production must be heat until 900°C with exclusion of air in coke oven. This process condition may lead to higher cost operating. The other condition that leads to higher operating cost is this process needs at least 7 columns in the plant. This process also produced solid residue. This problem requires a larger area for disposal purpose. Human being will be affected by this process. Skin cancer, malignant skin tumor and scrotal cancer are the examples of problem that human will be faced when decided to have this process without any proper safety precaution or equipment. In conclusion, studies in experimental systems and in surrogate tissues of humans provide strong evidence for a genotoxic/mutagenic mechanism underlying the effects of occupational exposures during coal-tar distillation (Ullmann, 2005).

2.1.1.4 Toluene – Benzoic Acid Process About 5% of phenol is produced using the oxidation of toluene, a process which also known as the Dow and California Research Process. In this process, phenol is produced by undergoing two times oxidation and one hydrolysis process. Thus, it includes three main reaction before producing phenol as product. The first reaction is oxidation of toluene to form benzoic acid. Toluene will react with air that being supply to the reactor with the presence of catalyst. The catalyst used in this process is cobalt salt with concentration between 0.1% and 0.3%. The process undergo under temperature between 121˚C and 177˚C and the reactor where the process takes place is operated at pressure of 206 kPa. The reactor effluent will be distilled and the purified benzoic acid is collected for further process. The overall yield of this process is believed to be about 68 mol% of toluene (Kirk-Othmer). The second reaction is the oxidation of benzoic acid and being hydrolyzed into phenol that carried out in two reactors that are in series. In the first reactor, oxidation of benzoic acid produced phenyl benzoates also with presence of catalyst. The catalyst used in this process is mixture of copper and magnesium salts. The reactor operates at temperature of 243˚C and pressure of 147 kPa. The phenyl benzoates will then transferred to the second reactor and hydrolysed with steam to produced phenol and carbon dioxide as by-product. The process occurs at 200˚C and at atmospheric pressure and the overall yield of phenol from benzoic acid is around 88 mol% (Kirk-Othmer).

14

PLANT DESIGN FOR PRODUCTION OF PHENOL The reaction equations of this process are as follows: O CH3 + KMnO4

2

COH ‖

O

O

COH + 1/2O2 ‖

C‖

+ H2O + CO2

O

O C‖

O O

+ H2O

OH +

COH ‖

The price of raw material known as toluene is 7162.50 RMB which equals to 4,316.77 Malaysian Ringgit/1040.78 US Dollar per metric tonne. The focusing price in this process is the price of phenol which China price list have stated on 10th October 2018 to be 11937.50 RMB that is equal to 7,194.61 Malaysian Ringgit/1734.64 US Dollar (Industrial Sector, 2018). Thus, the economic potential of this process is said to be 2877.84 Malaysian Ringgit/693.86 US Dollar. However, this process also has its drawback especially regarding catalyst. Catalyst recovery can be another problem for homogenous catalytic systems. Heterogeneous catalysts for selective oxidation of toluene in liquid phase have been proposed and may be employed. However, this form of catalysis also has disadvantages because the industrial waste generated involves solids containing transition metals of environmental concern, such as copper, 5 manga-nite and chromite. Recently, the use of Au–Pd alloy nanoparticles as highly active catalysts was reported for the oxidation of the carbon-hydrogen alkyl groups in toluene at 160˚C and under 10 bar with the main product generated being benzyl benzoate (Sarina, Huaiyong, Zhanfeng, & Guoran, 2012). Other case, this process is high energy-consumed. Especially in present times, it will consumed up to 3 to 4 times more energy compared to oxidation of cumene process. Thus, it is not widely used in present industrial activities to produce phenol.

15


2.2

Profit Margin

2.2.1 Phenols from Toluene Benzoic Acid From process of Toluene-Benzoic Acid Process in producing cumene, the following prices for raw material(s) and product(s) are found: Toluene

= 4,316.77 Malaysian Ringgit per tonne = 1040.78 US Dollar per tonne

Phenol

= 7,194.61 Malaysian Ringgit per tonne = 1734.64 US Dollar per tonne

Molecular weight (MW) of Toluene is 92.14 kg/kmol and Phenol is 100.15 kg/kmol. By using 100 kmol of toluene as feed basis,

Cost of Raw Material: Cost of Raw Material = Feed Basis x MW x Cost of Material per tonne Cost of Toluene = 100 kmol(92.14

kg RM4316.77 1 tonne )( )( ) kmol tonne 1000 kg

𝐂𝐨𝐬𝐭 𝐨𝐟 𝐓𝐨𝐥𝐮𝐞𝐧𝐞 = 𝐑𝐌 𝟑𝟗, 𝟕𝟕𝟒. 𝟕𝟐 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐓𝐨𝐥𝐮𝐞𝐧𝐞 = $ 𝟗, 𝟒𝟗𝟒. 𝟔𝟐

Cost of Product: Cost of Product = Feed Basis x MW x Cost of Material per tonne Cost of Phenol = 100 kmol(100.15


𝐂𝐨𝐬𝐭 𝐨𝐟 𝐏𝐡𝐞𝐧𝐨𝐥 = 𝐑𝐌 𝟕𝟐, 𝟎𝟓𝟒. 𝟎𝟐 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐏𝐡𝐞𝐧𝐨𝐥 = $ 𝟏𝟕, 𝟐𝟎𝟎. 𝟎𝟐

In order to find the profit margin, equation below is used: Profit Margin = Total Cost of Product − Total Cost of Raw Material Profit Margin = RM 72,054.02 − RM 39,774.72 = 𝐑𝐌 𝟑𝟐, 𝟐𝟕𝟗. 𝟗𝟑 Profit Margin = $ 17,200.02 − $ 9,494.62 = $ 𝟕, 𝟕𝟎𝟓. 𝟒𝟎

16

PLANT DESIGN FOR PRODUCTION OF PHENOL 2.2.2 Phenols From Coal The following prices for raw material(s) and product(s) are found Raw Material 1) Coal

= RM 414.94/ton as to May 2018

(Coal, Australian thermal coal monthly price-Malaysian Ringgit per Metric Ton) 2) Cresol

= USD 2593/ton = RM 10 864.67/ton

(Cresol Price-China Cresol Price Manufacturers&Suppliers|made in China) 3) Xylenols

= USD 8000/ton =RM 33 520/ton

4) Phenols

= RM 7 194.61/ton

(Sunsirs Commodity Data Group, 2018)

Molecular weight (MW) of Coal, Cresol, Xylenols, and Phenols are 92.14 kg/kmol and Phenol are 207.252 kg/kmol, 108.14 kg/kmol, 122.167 kg/kmol and 100.15 kg/kmol, respectively. By using 100 kmol of coal as feed basis,

Cost of Raw Material:

Cost of Raw Material = Feed Basis x MW x Cost of Material per tonne Cost of Coal = 100 kmol(207.252


𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐨𝐚𝐥 = $ 𝟐 𝟎𝟓𝟏. 𝟒𝟔 = 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐨𝐚𝐥 = 𝐑𝐌 8 599.71

Cost of Product:

Cost of Product = Feed Basis x MW x Cost of Material per tonne Cost of Phenol = 100 kmol(100.15


𝐂𝐨𝐬𝐭 𝐨𝐟 𝐏𝐡𝐞𝐧𝐨𝐥 = 𝐑𝐌 𝟕𝟐, 𝟎𝟓𝟒. 𝟎𝟐 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐏𝐡𝐞𝐧𝐨𝐥 = $ 𝟏𝟕 𝟐𝟎𝟎. 𝟎𝟐 Cost of Product = Feed Basis x MW x Cost of Material per tonne Cost of Cresols = 100 kmol(108.14 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐫𝐞𝐬𝐨𝐥𝐬 = 𝐑𝐌 𝟏𝟏𝟕 𝟒𝟗𝟎. 𝟓𝟒 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐫𝐞𝐬𝐨𝐥𝐬 = $ 𝟐𝟖 𝟎𝟐𝟕. 𝟑𝟕

kg RM10 864.67 1 tonne )( )( ) kmol tonne 1000 kg

17

PLANT DESIGN FOR PRODUCTION OF PHENOL Cost of Xylenols = Feed Basis x MW x Cost of Material per tonne Cost of Xylenols = 100 kmol(122.167

kg RM 33 520 1 tonne )( )( ) kmol tonne 1000 kg

𝐂𝐨𝐬𝐭 𝐨𝐟 𝐗𝐲𝐥𝐞𝐧𝐨𝐥𝐬 = 𝐑𝐌 𝟒𝟎𝟗 𝟓𝟎𝟑. 𝟕𝟑 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐗𝐲𝐥𝐞𝐧𝐨𝐥𝐬 = $ 𝟗𝟕 𝟔𝟖𝟕. 𝟏𝟏

In order to find the profit margin, equation below is used: Profit Margin = Total Cost of Product − Total Cost of Raw Material Profit Margin = RM 599 048.29 − RM 8 599.71 = 𝐑𝐌 𝟓𝟗𝟎 𝟒𝟒𝟖. 𝟓𝟖 Profit Margin = $ 17,200.02 − $ 9,494.62 = $ 𝟏𝟒𝟎 𝟖𝟓𝟏. 𝟓𝟏

2.2.3 Phenol From Cumene Peroxidation From process of production of phenol from cumene the following are the prices for raw material(s) and product(s): Cumene

= 6,093.39 Malaysian Ringgit per tonne = 1,455 US Dollar per tonne

Phenol

= 7,194.61 Malaysian Ringgit per tonne = 1,734.64 US Dollar per tonne

Acetone

= 14,238.86 Malaysian Ringgit per tonne = 3,400 US Dollar per tonne

Crude Alpha Methyl Styrene = 6,072.45 Malaysian Ringgit per tonne = 1,450 US Dollar per tonne Tar

= 1,612.34 Malaysian Ringgit per tonne = 385 US Dollar per tonne

Molecular weight (MW) of Cumene is 120.19 kg/kmol, Phenol is 100.15 kg/kmol, Acetone is 58.08kg/kmol, Crude Alpha Methyl Styrene is 118.18kg/kmol and Tar is 15,000kg/kmol. By using 100 kmol of cumene as feed basis,

18

PLANT DESIGN FOR PRODUCTION OF PHENOL Cost of Raw Material: Cost of Raw Material = Feed Basis x MW x Cost of Material per tonne Cost of Cumene = 100 kmol(120.19


𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐮𝐦𝐞𝐧𝐞 = 𝐑𝐌 𝟕𝟑, 𝟐𝟑𝟔. 𝟒𝟓 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐮𝐦𝐞𝐧𝐞 = $ 𝟏𝟕, 𝟒𝟖𝟕. 𝟔𝟑

Cost of Product: Cost of Phenol = 100 kmol(100.15


𝐂𝐨𝐬𝐭 𝐨𝐟 𝐏𝐡𝐞𝐧𝐨𝐥 = 𝐑𝐌 𝟕𝟐, 𝟎𝟓𝟒. 𝟎𝟐 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐏𝐡𝐞𝐧𝐨𝐥 = $ 𝟏𝟕, 𝟐𝟎𝟎. 𝟎𝟐

Cost of Acetone = 100 kmol(58.08


𝐂𝐨𝐬𝐭 𝐨𝐟 𝐀𝐜𝐞𝐭𝐨𝐧𝐞 = 𝐑𝐌 𝟖𝟐, 𝟔𝟗𝟗. 𝟑𝟎 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐀𝐜𝐞𝐭𝐨𝐧𝐞 = $ 𝟏𝟗, 𝟕𝟒𝟕. 𝟐𝟎

Cost of Crude Alpha Methyl Styrene = 100 kmol(118.18

kg RM6,072.45 1 tonne )( )( ) kmol tonne 1000 kg

𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐫𝐮𝐝𝐞 𝐀𝐥𝐩𝐡𝐚 𝐌𝐞𝐭𝐡𝐲𝐥 𝐒𝐭𝐲𝐫𝐞𝐧𝐞 = 𝐑𝐌 𝟕𝟏, 𝟕𝟔𝟒. 𝟐𝟏 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐂𝐫𝐮𝐝𝐞 𝐀𝐥𝐩𝐡𝐚 𝐌𝐞𝐭𝐡𝐲𝐥 𝐒𝐭𝐲𝐫𝐞𝐧𝐞 = $ 𝟏𝟕, 𝟏𝟑𝟔. 𝟎𝟖

Cost of Tar = 100 kmol(15,000

kg RM1,612.34 1 tonne )( )( ) kmol tonne 1000 kg

𝐂𝐨𝐬𝐭 𝐨𝐟 𝐓𝐚𝐫 = 𝐑𝐌 𝟐, 𝟒𝟏𝟖, 𝟓𝟏𝟎 𝐂𝐨𝐬𝐭 𝐨𝐟 𝐓𝐚𝐫 = $ 𝟓𝟕𝟕, 𝟒𝟗𝟗. 𝟒𝟔

In order to find the profit margin, equation below is used: Profit Margin = Total Cost of Product − Total Cost of Raw Material Profit Margin = (RM 72,054.02 + RM 82,699.30 + RM 71,764.21 + RM 2,418,510) − RM 73,236.45 Profit Margin = 𝐑𝐌 𝟐, 𝟓𝟕𝟏, 𝟕𝟗𝟏. 𝟎𝟖 Profit Margin = $ 631,588.03 − $ 1,455 = $ 𝟔𝟑𝟎, 𝟏𝟑𝟑. 𝟎𝟑

19


3.0

SITE SELECTION

There are 3 places that have been decided which are Pasir Gudang, Kerteh and Gebeng. The place chosen based on several general descriptions. The general description of places is as follow:1. Market Availability The market must be available to ensure the product can be sold to generate profit. The proximity to the major markets is vital for the site selection. It give an advantages to the customer to purchase from nearby sources, thus the location needs to be near the market demand as to reduce the product cost distribution and time required for shipping. The side – product which is acetone also need to be considered for the market since this chemical also plays an important part in the chemical industries.

2. Transportation Water, railroads and highways are the common means of transportation used by major industrial concerns. The local freight rates and existing railroad lines should be considered. The proximity to railroad centers and the possibility of ocean transport must be considered. The kind and amount of products and raw materials determine the most suitable type of transportation facilities.

3. Availability of labor Labour supply is needed for construction of the plant and its operation. There should be an adequate pool of skilled and unskilled labour locally and the labour is suitable to be trained for enhancement of their skill. The prevailing pay scales, restrictions on number of hours worked per week should also be considered.

4. Utilities and Facilities Utilities Supply is another crucial requirement in selecting the site for plant. Utilities supplies includes power supply and water supply. Power supply will be received from the nearest Tenaga Nasional Berhad (TNB) as TNB is the main electricity power generator and supplier in Peninsular Malaysia.

20

PLANT DESIGN FOR PRODUCTION OF PHENOL 5. Land Estimation Land and soil structure should be examined carefully before selecting the site. The difference in soil structure should be study earlier because it can be divided into three types of soil which are loam-type, laterite-type and sandy-type. Topography of land must also be considered as it will affect the construction cost. This is because if a plant is built on a hilly area, there will be an extra cost to levelling the hills. Thus, a study on the topography of the site must be inspected as well and a satisfying area which is a flat area, should be first to be chose as the site of the plant. Additional space is required for a preparation in future changes. This is due to the company’s target which is after 10 years of production, scaling up will be performed to increase the methanol production in order to meet the customer needs. The cost of the plant is also important as well as local building costs and living condition.

6. Climate Climate is also one of the important selectivity to be brought into account. The extreme condition of a place is required for consideration since excessive humidity or extremes of hot or cold weather will induce a need for the plant to have a special consideration on the plant equipment.

7. Waste Management Methods of disposing waste materials from the process industries must obliges the legal restrictions. The selected site should have adequate capacity and facilities for correct waste disposal. The area should has minimal restrictions on pollution. The permissible tolerance levels for various methods of waste disposal and potential requirements for additional waste treatment facilities should be considered.

8. Raw Material Availability The raw materials availability is an important aspect in the site selection consideration. Without raw material, there will be no product. Thus, the associates’ factor of the raw material resources should be determined at the considered location. The location must be closed with the raw materials source in order to reduce the transportation and storage charges. The price of the raw materials must be low and the raw material must be pure. The storage requirements should also be considered for this part.

21

PLANT DESIGN FOR PRODUCTION OF PHENOL Table 4 Shows The Criteria Of The Site Selection Site Suggestion No

Selection Criteria

Kerteh,

Pasir

Gebeng,

Terengganu

Gudang,

Pahang

Johor 1

Raw Material Availability

3

3

3

2

Market Availability

5

5

4

3

Energy Availability and

4

3

4

Water Supply 4

Climate Conditions

3

4

3

5

Transportation facilities

5

5

5

6

Land Estimation

4

4

3

7

Waste Management

3

5

4

8

Labour Supply

4

4

3

31

33

29

Total

1 = Very Bad , 2 = Bad , 3 = Moderate , 4 = Good , 5 = Very Good Based on the market survey and preliminary feasibility study, there are few places have been considered for the site selection. Three sites within Peninsular Malaysia were selected for further consideration which are Kerteh, Pasir Gudang and Gebeng. The location chosen is Pasir Gudang, Johor.

Raw Material Availability For all three chosen site there is no raw material availability which is cumene. Thus, the required raw material is to be imported from overseas market which is India, and Japan.

22


3.1

Kerteh, Terengganu

3.1.1 Market Availability

Figure 12 Based on “Statistical Review of World Energy 2008” Based on the figure above, at Kerteh there is no production of phenol. Kerteh is one of chemical industries known as ethylene based production. The advantages of our company to open at Kerteh are high due to no competition from the other company at Kerteh. The demands for phenol exactly high from years to years due to highly produce phenol based product. 3.1.2 Transportation 3.1.2.1 Air It is convenient not only for business and leisure travelers but also for transporting cargos. Terengganu’s main air hub, Sultan Mahmud Airport in Kuala Terengganu has new terminal building and runway that is constructed to cater passengers and cargo traffic. Thus, if we want to purchase equipment or machinery, we can use the air service as the airport flies about half a dozen flights per day. It is easy for people working in our industry to travel in and out of Kerteh either for business or personal purpose. For instance, when there is an urgency for meeting somewhere out of Kerteh, it is easy for us to travel.

23


Figure 13 Sultan Mahmud Airport in Kuala Terengganu

3.1.2.2 Roads and Highways Terengganu has invested billions in a well-connected network of roads and highways which links major towns, seaports, airports and industrial centers in east, west, south and north Peninsular Malaysia. Currently, Terengganu has about 1,071 km federal roads and about 1,660 km state roads. It is targeted to complete the phase 2 of East Coast Highway which links Kuala Lumpur and Kuala Terengganu.

Interstate travel and commerce has also risen with the opening of the Simpang PulaiLojing, Kuala Berang road that links Hulu Terengganu district to Grik in Perak, West Malaysia which slashes travel time from East Coast to West Cost of Peninsula down to 4 hours instead of 10 hours.

24

PLANT DESIGN FOR PRODUCTION OF PHENOL Roads and highways in Terengganu make it easy for transportation of materials, equipment and machinery from Kuala Lumpur for instance. Transportation such as cars, lorries, tankers can be used to transport our raw materials and products to West Coast of Peninsular Malaysia which are more cheaper compared to air transportation.

Figure 14 shows East Coast Expressway (Phase 2)

25

PLANT DESIGN FOR PRODUCTION OF PHENOL Table 5 Shows The Container Haulage Rates (The Hauliers) State Kerteh, Terengganu Haulage Rates 20’ & 40’

RM2,782.00

Road Tolls

RM197.80

FAF (16% Period from 1st June 2011)

RM450.41

Total

RM3430.21/ US$803.33

3.1.2.3 Railway A 77 km dedicated industrial railroad from Kerteh to Kuantan serves companies in petrochemical corridor. Built and operate by Petronas, this railroad helps in transporting cargo back and forth between key seaports in Kuantan and Kerteh. Extensions are planned for Paka and Kemaman.

The existence of railway also makes it easy for us to either import or export raw materials or products for our industry whereby such option, gives us the opportunity to choose which transportation is better and easier. It is also can be used by people working in our industry to travel back and forth.

Figure 15 shows East Coast Rail Line (ECRL)

26

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.1.3 Availability Of Labor There are few programs that has been introduced by the Malaysian government. One of the programs include a training centre which is Institut Teknologi Petroleum Petronas (INSTEP). There are three main areas involved at INSTEP which are: 1) Bridging Programmes – A program that has been developed to produce job-ready technicians and operators, as well as advanced programmes for engineers. 2) Professional Development Programmes- Developed to upskill working technicians and operators, as well as engineers. 3) Competency Based Assessment and Certification- Developed for industry workforce in meeting statutory requirements.

This INSTEP program helps Malaysian industries to have continuous workforce and to have employees whom are ready for challenges in industries. It also makes it easy for our neither industry to look for new employees whether engineers nor technicians. With this training centre, it makes it easy for us to direct employees and time saving for us where it reduces our workload of training untrained employees.

Figure 16 shows statistics for employment in Malaysia from 2007 to 2017 (Source: Statista) Referred to the figure, it shows that there is always and employment for industries. However, the employment availability surely depends on states and our country’s economics besides demands of our products. In every 5 years, the economic cycle will keep repeating but we can never know what challenges await upon us.

27

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.1.4 Utilities and Facilities Infrastructural facilities such as road accesses, electricity supply from the TNB, water supply, gas supply and telecommunications are readily available at Kerteh, Terengganu. There also have facilities such as banking, hospital and fresh water supply. The water supply falls under the jurisdiction of the state, and is treated and distributed by the Terengganu Water Company (SATU). Industrial sectors are charge RM1.15 per m3 with a rm50.00 minimum charge.

Figure 17 Shows The Tariff Of Water For Terengganu.

28

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.1.5 Land Estimation The land price is around RM0.18 – RM5.60 which is the cheapest compared to other states in Malaysia. Where land sells from as low as RM2.00 – RM4.50 per square feet to highs of RM18.00 – RM22.00 per square feet.

Figure 18 shows the Terengganu industrial area.

29

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.1.6 Climate

Figure 19 Shows Average Temperature And Precipitation (Source: Meteoblue) Referred to the figure, the mean daily maximum (solid red line) shows maximum temperature of an average day for every month in Kerteh. For mean daily minimum (solid blue line) shows average minimum temperature. Hot and cold nights (dashed red and blue lines) shows average hottest and coldest day and night for each moth of the last 30 years. Precipitation chart is used in identifying seasonal effects such as monsoon climate or wet season. Monthly precipitations above 150 mm are considered wet while below 30 mm are considered dry.

3.1.7 Waste Management Kerteh employs an integrated environmentally friendly waste-management from the industrial waste. There are two types of waste management (Packaged waste or Bulk Waste). Scheduled waste management is provided by Aldwech-Enviro Management At Teluk Kalong Industrial Estate. Table 6 Shows The Packaged Waste Price Packaged Waste 1. Pumpable Liquid per tonne

From 810.00 – RM3,150.00 Depending from the type of the waste.

2. Solid Waste per tonne

From 2,790.00 – RM 3,600.00

30

PLANT DESIGN FOR PRODUCTION OF PHENOL Table 7 Shows The Bulk Waste Price Bulk Waste 1. Pumpable Liquid per tonne

RM630.00 – RM1,800.00

2. Solid Waste per tonne

RM2,700.00

Sewage services are carried by Indah Water Consortium. Charges are based on number of employee.

Table 8 Shows The Category Of Price Rate Category

Rate based on number of employees

Premises receiving Individual Sepic Tanks

RM2.00 per head per month

services Premises with Connected Sewerage Services

RM2.50 per head per month

Table 9 Shows Transportation Rates For Waste (KualitiAlam, n.d.) State

Per pallet (min 18 pallet)

Kerteh, Terengganu 1st tier

RM180.17/US$42.19

2nd tier

RM90.09/US$21.10

Per metric tonne

RM225.22/ US$52.74

31


3.2

Gebeng, Pahang

3.2.1 Market Availability Table 10 List of Products Manufactured in Gebeng. Petrochemicals Zone

Facilities & Infrastructure

Products • Acrylic Acid and Esters • Syngas • Butyl Acrylate • Oxo-alcohols • Phthalic Anhydride and Plasticizers • Butanediol

• Peninsular Gas Utilisation (PGU) project Gebeng, Pahang

• Centralized utility facilities • Kuantan Port • Environment Technology Park • East Coast Highway

• Tetrahydrofurane • Gamma-butyrolastone • Polyester Copolymers • Purified Terephthalic Acid • Dispersion Polyvinyl Chloride • Methyl Methacrylates Copolymers • MTBE • Propylene • Polyacetals • Polypropylene • Polybutylene Terephthalate (PBT)

Based on Table 10, it shows “Statistical Review of World Energy 2008” for Gebeng Industrial Estate. The area indicates that there is no production of phenol at the zone area, which it gives some advantages to manufactures due to less competition. However, it finds disadvantageous to purchase the raw material from nearby sources. These will affect the high of production or distribution cost, and also the time required for shipping.

32

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.2.2 Transportation The location of the plant is at Gebeng. Pahang is the convenient and efficient because easy access due to its closeness to the international port which is Kuantan port. Our product can be exported easily to the international market because the facility at Kuantan port is very good. Besides that, our product can be selling in local market easily because there are highways which connect the east peninsula Malaysia to west peninsula Malaysia. For example Jalan Pintasan Kuantan highway and PLUS highway are the examples for road transportation.

Table 11 Shows The Container Haulage Rates (The Hauliers) State Kuantan, Pahang Haulage Rates 20’ & 40’

RM2,328.00

Road Tolls

RM197.80


RM376.90

Total

RM2902.70/ US$679.79

*FAF = Fuel Adjustment Factor

3.2.3 Availability Of Labor Plant location must require skilled labors. The labors and skilled workers are easily found in Gebeng Industrial Estate. It is because the location is near to the community and many factories around there so one of the place to find job. In Gebeng, there are also can provide intern and workers from local university or international university such as UiTM Jengka students, University Malaysia Pahang students and other.

3.2.4 Utilities and Facilities The Gebeng Industrial Estate (GIE) obtains its power supply from the Tanjung Gelang substation which generates power at a capacity of 2x15 MVA. Power supply is available via the exiting TNB substation (intake station) located within the GIE less than 3 km south to the south of the project site. The main electricity supplier is Tenaga Nasional Berhad (TNB). The electricity supply provided by TNB for Phase I and II is supported by its 132/11kV main intake and for Phase III, two sources of electricity supply are available which are Centralised Utility Facilities (CUF) and 12/275 kV main intake.

33

PLANT DESIGN FOR PRODUCTION OF PHENOL The water supply is supplied from the Semambu Water Treatment Plant. The current capacity of water supply is 2MG/D. The Government of Pahang is committed in ensuring an efficient water supply in Gebeng. The steps that have been taken are increase of water supply to 64 MG/D, building of a new 200 acre dam in Sungai Lembing, Kuantan and building of new pipes and water tanks in Gebeng Industrial Estate. Malaysia has one of the largest natural gas reserves in the Asia Pacific region; an estimated 85 trillion cubic feet. Currently, the natural gas suppliers for Gebeng Industrial Estate are Gas Malaysia and Petronas Gas Berhad. They supply industrial gases within the estate to cater for the overwhelming demand from existing and future petrochemical projects in the area.

The main telecommunication supplier is Telekom Malaysia. They provide state-of-theart facilities and services such as Integrated Systems Digital Network (ISDN), digital line, MAYPAC, Internet and video conferencing. Accessibility to Gebeng Industrial Estate is excellent. The East Coast Highway which links Kuantan and Kuala Lumpur has an estimated travel time of just 2 hours. In addition, a railway link connects the integrated petrochemical complex in Kerteh (Terengganu) to Gebeng and Kuantan Port. This railway link further strengthen the chemical and petrochemical linkage between Gebeng and other industrial centres thus ensures safe transport as compared to transportation of dangerous goods by road.

The Pahang Fire and Rescue Department is located near Gebeng in order to handle any emergencies. In addition, located within the vicinity of Gebeng is Petronas Centralised Emergency Facilities. Both stations are equipped with HAZMAT (hazardous material) facilities. Another services provided is the Gebeng Emergency Mutual Aid (GEMA), a voluntary crisis management organisation, set up from an alliance between Government agencies and private manufacturers in Gebeng. The main objective of GEMA is to execute proactive action and they also offer expert services to counter emergencies.

3.2.5 Land Estimation The land price at Gebeng range is between RM 14.00 - RM 23.00 per square feet. The possibility to occur flooding disaster is low because Gebeng is a hilly location. Besides that, sufficient and suitable land must be available for the proposed plant and for future expansion. Safety precaution must be taken when building plants near the ocean to make sure whether it has the possibility to cause the tsunami disaster. Based on Figure 21, Gebeng have many

34

PLANT DESIGN FOR PRODUCTION OF PHENOL petrochemical industry and moreover, the propose plant far from the community. It can affect the community because some of the industry will produce hazardous substance to surroundings.

Figure 20 Gebeng Industrial Estate Map

Figure 21 The Location Of Propose Plant From Port Kuantan

35

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.2.6 Climate Climate and environment affect the operational efficiency of workers which excess cold may cause tiredness, fatigue or disease among workers. Decreased humidity and climate temperature high can cause dehydration to the employees while working. Therefore, the climate in Gebeng Industrial Estate are hot and wet all year which the temperature range is 30°C to 32°C and have the northeast monsoon that supplied humidity in that area.

3.2.7 Waste Management The primary source of sewage and waste are sanitary waste, process drains and surface drainage. The sewage system designed separately to conduct these to disposal without logged with solid or filled with dangerous concentration of explosive. Table 12 Shows The Transportation Rates For Waste (KualitiAlam, n.d.) State


Gebeng, Pahang 1st tier

RM84.21/US$19.72

2nd tier

RM42.11/US$9.86

Per metric tonne

RM105.26/ US$24.65

36


3.3

Pasir Gudang Industrial Estate, Johor Bahru

3.3.1 Market Availability The product of the plant is phenol and acetone. However the company is aiming for phenol as the main product. Phenol is widely used in industry and the precursor to many materials and useful compounds. The global phenol market predicts to witness a robust CAGR of 6.8% from 2017 to 2022. The cumene process segment should be worth almost US$12.3 billion in 2017. Table 13 Shows A Few Prospect Market Available In Pasir Gudang Product Produced From Phenol

Company

Paint

Nippon Paint (M) Sdn Bhd 23, Jalan Bacang 36, Taman Kota Masai, 81700 Pasir Gudang.

Further Processed Food

PK Agro Industrial Products Sdn Bhd 100, Jalan Nibong 3, Kawasan Perindustrian Tanjung Langsat, Pasir Gudang.

Protective Coating

TK Blasting & Coating Sdn Bhd PTD 662, Jalan Platinum Utama, 81700 Pasir Gudang, Johor.

Epoxy coating

EEW Malaysia Sdn Bhd PLO 109, Jalan Tengar, Tanjung Langsat Industrial Complex, 81700 Pasir Gudang.

3.3.2 Transportation Pasir Gudang Industrial Estate is strategic as it is connected to Johor Bahru through a four-lane Lebuhraya Pasir Gudang, JB Eastcoast Highway and a trunk road. There are also several main roads connected from this area to Pasir Gudang city centre which ease the transportation network.

37

PLANT DESIGN FOR PRODUCTION OF PHENOL Table 14 Shows The Container Haulage Rates (The Hauliers) State Pasir Gudang Industrial Estate, Johor Bahru Haulage Rates 20’ & 40’

RM2,818.00

Road Tolls

RM211.00


RM456.23

Total

RM3485.23/ US$816.21

*FAF = Fuel Adjustment Factor 3.3.3 Availability Of Labor Pasir Gudang is currently rising in terms of the industrial industry. So, people migrates to the city for the job opportunities. Johor Skills Hub is one of the platform to develop suitably skilled workforce for Johor in particular. According to (Payscale, 2018), the average salary is RM46 378. Mechanical engineer, Account Executive and Project Engineer range of salary between RM 39 076 and RM 50 330 per year. A single operator is believed to work for 5 days per week with each day of 8 hour shifts a week. An analysis from MIDA, unskilled employee range salary from RM1,253/month until RM3,683/month for non – executive position. In addition, the average monthly salary for executives, senior manager and top executives are RM3,779, RM15,614 and RM32,043, respectively. (MIDA, 2017)

3.3.4 Utilities and Facilities The power is supply through 500kVcable running at 275kV which is connected to the grid. While water supply will be received from Syarikat Air Johor Sdn.Bhd (SAJ Holdings) as it is the department that have been obligated to serve water supply to the communities and industries within Pasir Gudang area. Table 15 Shows The Water Supply Prices In Johor State

Per m3

Pasir Gudang Industrial Estate, Johor Bahru 0 – 35 m3

RM2.80

More than 35 m3

RM3.30

Minimum charge

RM30.00

38

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.3.5 Land Estimation The most suitable soil for site of the new chemical plant is a loam-type soil with a big ratio of hard rock in the layers of the ground. Natural soils in the site area include acidic soils of Kranji series which consist of saline gley soil and acid sulphate soils. This site also satisfies the consideration for a flat area instead of hilly area. For heavy industries the cost for land is RM 2400/hectar with annual assessment rate of 0.33% up to 1% of property value (MIDA States).

Figure 22 shows the Pasir Gudang Industrial Estate Map

39

PLANT DESIGN FOR PRODUCTION OF PHENOL 3.3.6 Climate The climatic information of Pasir Gudang area is based from data obtained from Meteorological Station, Senai. From the data, this area is likely to be influenced by two different monsoonal seasons which are Northeast Monsoon and Southwest Monsoon which are likely occurred from March to April and September to October respectively. It is expected that high rainfall with high relative humidity will hit this area with average annual rainfall of 2498.9 mm and 24 hours mean relative humidity from 83.2% to 87.8%. The mean annual temperature would likely to be 26.1˚C.

3.3.7 Waste Management Acid waste without chromate is at RM1440 per 800 litre pallet tank per tonned or US$337 per 800 litre pallet tank per pallet. For 200 L drum per tonne is at RM1620 and per pallet is at US$379. Table 2 shows the transportation rates for waste for the respective state Table 16 Transportation Rates for Waste (Source: (KualitiAlam, n.d.)) State


Pasir Gudang Industrial Estate, Johor Bahru 1st tier

RM82.25/US$19.26

2nd tier

RM41.13/US$9.63

Per metric tonne

RM102.82/ US$24.08

40


BLOCK FLOW DIAGRAM FOR PHENOL PLANT H2 O

Vent Gases NaOH

H2 SO4

Aqueous Stream

Organic Stream

Cumene Phase Separator (S1)

Pre – Purification Reactor (R3)

Acidified Wash Water

𝐏𝐡𝐞𝐧𝐨𝐥 𝟗𝟔% 𝐏𝐮𝐫𝐢𝐭𝐲

Distillation Unit 1

H2

Crude Phenol

Scrubber (S2)

Distillation Unit 2

Cleavage Unit (R2)

Distillation Unit 3

Oxidation Reactor (R1)

Distillation Unit 5

Air

Distillation Unit 4

4.0

41


5.0

PROCESS FLOW DIAGRAM FOR PHENOL PLANT

42


43

Table 17 Shows The No Of Stream, Temperature, Pressure, Molar Flow Rate and Heat Flow For Phenol Production Plant Stream

1

2

3

4

5

6

7

8

Temperature (°C)

25

25

-

100

100

-

110

25

Pressure (atm)

6

6

-

6

6

-

6

6

100

1904.76

-

100

719

-

100

100

0

0

-

358888951.1

3610304.4

-

151857055.6

0

Molar Flow rate (Kmol/h) Heat flow (kJ/h)


9

10

11

12

13

14

15

16

Temperature (°C)

25

-

100

-

100

100

175

152

Pressure (atm)

6

-

1

-

1

1

1

1

Molar Flow rate (Kmol/h)

0

-

100

-

72.41

28.14

18

17.87

Heat flow (kJ/h)

0

-

136337110

-

99225378

83369109.6

1359110318.3 415108.87


17

18

19

20

21

22

23

Temperature (°C)

152

100

25

25

175

180

180

1

1

1

1

1

1

1

3.06

5.36

26.16

100

53.84

0.51

5.36

13245557.8

508273.85

0

0

172595120.2

169324023.7

9024639.35

Pressure (atm) Molar Flow rate (Kmol/h) Heat flow (kJ/h)


6.0

MATERIALS BALANCE

6.1

Reactor 1 – Oxidation Reactor Vent Gases n4 O2

NaOH

n5 N2

n6 CH3 OH n1 C6 H5 CH(CH3 )2

n2

kmol x1 = 0.21 O kmol 2

n3

kmol x2 = 0.79 N kmol 2

Oxidation Reactor (R1)

n7 C6 H5 CH(CH3 )2 n8 C6 H5 COOH(CH3 )2 n9 C6 H5 COH(CH3 )2 n10 C6 H5 COCH3

Chemical Reaction Involved In The Oxidation Reactor Sodium hydroxide, NaOH acts as the emulsification agent for the emulsification of cumene in the reactor and not directly involve in the reaction of the reactants to produce the products. Thus, is not considered for the material balance. Emulsification increase the interfacial area to enhance the rate of reaction. After reaction, the vent gases is recycled back and the product is sent to the second reactor which is the cleavage unit.

Main Reaction 1) C6 H5 CH(CH3 )2 (ℓ) + O2 (g) → C6 H5 COOH(CH3 )2 (ℓ)____________________ξ1 Cumene

+

Oxygen →

Cumene Peroxide

Side Reactions 1) 2C6 H5 CH(CH3 )2 (ℓ) + O2 (g) → 2C6 H5 COH(CH3 )2 (ℓ)____________________ξ2 2 Cumene

+

Oxygen → a,a – dimethylbenzyl alcohol (DMBA)

2) C6 H5 COOH(CH3 )2 (ℓ) → C6 H5 COCH3 (ℓ) + CH3 OH (ℓ) __________________ξ3 Cumene Peroxide

→ Acetonephenone + Methanol

44

PLANT DESIGN FOR PRODUCTION OF PHENOL Assumptions 1) Ratio of feed; Cumene : Oxygen Gas = 1 : 4. 2) Fractional conversion of Cumene to Cumene Peroxide is 80%. 3) Fractional conversion of Cumene to DMBA is 6%. 4) Fractional conversion of Cumene Peroxide to Acetonephenonee and Methanol is 2.5%.

Material Balance Basis of Calculation = 100 kmol/h Cumene fresh feed Type of reaction

= Reactive reaction

Method

= Extend of Reaction, ξ

Input Output Component Molar Composition Component Molar Composition Flow Flow Rate Rate (kmol/hr) (kmol/hr) Gas Phase Gas Phase Oxygen 400 0.2100 Methanol 2 0.0011 (O2) (CH3OH) Nitrogen 1504.7619 0.7900 Oxygen 317 0.1738 (N2) (O2) Nitrogen 1504.7619 0.8251 (N2) Total 1904.7619 1.0000 Total 1823.7619 1.000 Liquid Phase Liquid Phase Cumene 100 1.0000 Cumene 14 0.1400 (C9H12) (C9H12) Cumene 78 0.7800 Peroxide (C9H12O2) DMBA 6 0.0600 (C9H12O) Acetonephenone 2 0.0200 (C8H8O) Total 100 1.0000 Total 100 1.0000

45


6.2

Reactor 2 - Cleavage Unit H2 SO4

n7 C6 H5 CH(CH3 )2 n8 C6 H5 COOH(CH3 )2

Cleavage Unit (R2)

n11 C6 H5 CH(CH3 )2 n12 C6 H5 COCH3

n9 C6 H5 COH(CH3 )2

n13 C6 H5 OH

n10 C6 H5 COCH3

n14 (CH3 )2 CO n15 C6 H5 C(CH2 )CH3 n16 H2 O H2 SO4

Chemical Reaction Involved In The Cleavage Unit Sulfuric Acid, H2SO4 (aq) is not directly involve in the reaction of the reactants to produce products. Thus, is not considered for the material balance. However, the aqueous acid solution is required to enable the hydrolysis of cumene peroxide.

Main Reaction 1) C6 H5 COOH(CH3 )2 (ℓ) → C6 H5 OH (g) + (CH3 )2 CO (g) Cumene Peroxide →

Phenol

+

Acetone

Side Reactions 2) C6 H5 COH(CH3 )2 (ℓ) → C6 H5 C(CH2 )CH3 (g) + H2 O (ℓ) DMBA

→ α – methylstyrene +

Water

Assumptions 1) Overall conversion of DMBA and Cumene Peroxide are 100%. 2) Main reaction: Selectivity of Phenol over Acetone is 65%. 3) Side reaction: Selectivity of α – methylstyrene over water is 87%. 4) Ratio of Sulfuric Acid. Cumenefeed at R1 : Sulphuric Acid = 100 : 0.7. Thus, molar flowrate of sulphuric acid, H2SO4 = 0.7 kmol/h.

46


47

Material Balance Type of reaction Component

Cumene (C9H12) Cumene Peroxide (C9H12O2) DMBA (C9H12O) Acetonephenone (C8H8O)

Total

= Non – reactive reaction Input Molar Composition Component Flow Rate (kmol/hr) 14.0000 0.1400 Cumene (C9H12) 78.0000 0.7800 Acetonephenone (C8H8O) 6.0000 2.0000

100.0000

0.0600 Phenol (C6H5OH) 0.0200 Acetone (C3H6O) α– methylstyrene (C9H10) Water (H2O) 1.0000 Total

Output Molar Composition Flow Rate (kmol/hr) 14.0000 0.1400 2.0000

0.0200

50.7000

0.5070

27.3000

0.2730

5.2200

0.0522

0.7800

0.0078

100.0000

1.0000

Molar flowrate of sulphuric acid, For every 100 kmol/h of cumene fed to reactor 1, 0.7 kmol/h of sulphuric acid is fed for reactor 2.


6.3

48

Separator 1 - Phase Separator (Gravity Settler)

The gravity settling yields to two streams which are phenol rich organic stream and acid rich aqueous stream. The aqueous stream is recycling back to the second reactor unit. Organic Stream n11 C6 H5 CH(CH3 )2 n12 C6 H5 COCH3 n11 C6 H5 CH(CH3 )2 n12 C6 H5 COCH3

n13 C6 H5 OH Phase Separator (S1)

n14 (CH3 )2 CO n15 C6 H5 C(CH2 )CH3

n13 C6 H5 OH

n16 H2 O

n14 (CH3 )2 CO

Aqueous Stream

n15 C6 H5 C(CH2 )CH3

H2 SO4

n16 H2 O

Traces of H2 O from n16

H2 SO4 Material Balance Component

Cumene (C9H12) Acetonephenone (C8H8O) Phenol (C6H5OH) Acetone (C3H6O) α– methylstyrene (C9H10) Water (H2O) Total

Input Molar Flow Rate (kmol/hr) 14.0000 2.0000 50.7000 27.3000 5.2200 0.7800 100.0000

Composition

Component

0.1400 Cumene (C9H12) 0.0200 Acetonephenone (C8H8O) 0.5070 Phenol (C6H5OH) 0.2730 Acetone (C3H6O) 0.0522 α – methylstyrene (C9H10) 0.0078 Water (H2O) 1.0000 Total

Output Molar Flow Rate (kmol/hr) 14.0000

Composition

0.1400

2.0000

0.0200

50.7000

0.5070

27.3000

0.2730

5.2200

0.0522

0.7800

0.0078

100.0000

1.0000


6.4

49

Separator 2 – Scrubber

Absorption technique is used with water as the solvent. The water extracts the remaining acids in the organic stream from phase separator and leaves as acidified wash water. The presence of the sulphuric acid even after going to gravity settler is due to the equilibrium separation factors for the sulphuric acid since although the acid has propensity to remain maximum in the aqueous phase, it also do have a little extent towards the organic phase. H2 O

Crude Phenol n11 C6 H5 CH(CH3 )2 n12 C6 H5 COCH3

n11 C6 H5 CH(CH3 )2

Scrubber (S2)

n12 C6 H5 COCH3

n13 C6 H5 OH

n13 C6 H5 OH

n14 (CH3 )2 CO

n14 (CH3 )2 CO



n16 H2 O

Acidified Wash Water

n16 H2 O H2 SO4 traces

Material Balance Component

Cumene (C9H12) Acetonephenone (C8H8O) Phenol (C6H5OH) Acetone (C3H6O) α– methylstyrene (C9H10) Water (H2O) Total

Input Molar Flow Rate (kmol/hr) 14.0000 2.0000 50.7000 27.3000 5.2200 0.7800 100.0000

Composition

Component

0.1400 Cumene (C9H12) 0.0200 Acetonephenone (C8H8O) 0.5070 Phenol (C6H5OH) 0.2730 Acetone (C3H6O) 0.0522 α– methylstyrene (C9H10) 0.0078 Water (H2O) 1.0000 Total

Output Molar Flow Rate (kmol/hr) 14.0000

Composition

0.1400

2.0000

0.0200

50.7000

0.5070

27.3000

0.2730

5.2200

0.0522

0.7800

0.0078

100.0000

1.0000


6.5

Distillation Unit

The crude phenol enters a series of distillation column to separate the products from the reaction. Since few component form azeotropes at atmospheric pressure when fractionated, vacuum condition is applied to the distillation units to bypass the formation of azeotropes. The pressure reduction enhances the relative volatility making it farther from unity and eliminates the formation of azeotrope. Thus, allowing purer products to be obtained.

n17 C6 H5 CH(CH3 )2

n21 C6 H5 CH(CH3 )2

n18 (CH3 )2 CO

n20 (CH3 )2 CO

n12 C6 H5 COCH3 n13 C6 H5 OH

n19 C6 H5 CH(CH3 )2 n20 (CH3 )2 CO n12 C6 H5 COCH3

n14 (CH3 )2 CO


Distillation Unit 3

n11 C6 H5 CH(CH3 )2

Distillation Unit 2

From Scrubber

Distillation Unit 1

n16 H2 O

n13 C6 H5 OH



n27 C6 H5 CH(CH3 )2

n16 H2 O


n23 C6 H5 CH(CH3 )2


n12 C6 H5 COCH3

n20 (CH3 )2 CO n30 C6 H5 C(CH2 )CH3

Distillation Unit 5

n23 C6 H5 CH(CH3 )2

n29 C6 H5 CH(CH3 )2

Distillation Unit 4

n33 C6 H5 OH

n13 C6 H5 OH

n25 C6 H5 CH(CH3 )2 n20 (CH3 )2 CO n26 C6 H5 C(CH2 )CH3

n24 C6 H5 C(CH2 )CH3 n35 C6 H5 C(CH2 )CH3 n12 C6 H5 COCH3

n36 C6 H5 OH

n31 C6 H5 CH(CH3 )2 n32 C6 H5 C(CH2 )CH3


Figure 23 Shows The Distillation Units For The Phenol Plant

50


51

6.5.1 Separator 3 – Distillation Unit 1 The first distillation unit separates mainly acetone from the other components. Inlet stream from scrubber which is the crude phenol.

Assumptions 1) Distillate composition: 98% of acetone from crude phenol, 100% of water from crude phenol, and 0.4% Cumene separated from the crude phenol. 2) Bottom composition: Remaining.


Cumene (C9H12) Acetonephenone (C8H8O) Phenol (C6H5OH) Acetone (C3H6O) α– methylstyrene (C9H10) Water (H2O)

Total

Input Molar Flow Rate (kmol/hr) 14.0000 2.0000

Composition

Component

0.1400

5.2200

0.0200 Acetone (C3H6O) 0.5070 Water (H2O) 0.2730 Cumene (C9H12) 0.0522 Total

0.7800

0.0078

50.7000 27.3000

100.0000

Cumene (C9H12) Acetone (C3H6O) Acetonephenone (C8H8O) Phenol (C6H5OH) α– methylstyrene (C9H10) 1.0000 Total

Output Molar Flow Rate (kmol/hr) Distillate

Composition

26.7540

0.9697

0.7800

0.0283

0.0560

0.0020

28.1360

1.0000

Bottom 13.9440

0.1926

0.5460

0.0075

2.0000

0.0276

50.7000

0.7002

5.2200

0.0721

72.4100

1.0000


52

6.5.2 Separator 4 – Distillation Unit 2 The second distillation unit separates mainly cumene and α – methylstyrene from the other components. Inlet stream from the bottom product of Distillation Unit 1.

Assumptions 1) Distillate composition: 99% of cumene from inlet stream, 70% of α – methylstyrene from inlet stream. 2) Bottom composition: Remaining.


Cumene (C9H12) Acetone (C3H6O) Acetonephenone (C8H8O) Phenol (C6H5OH) α– methylstyrene (C9H10)

Input Molar Flow Rate (kmol/hr) 13.9440 0.5460 2.0000 50.7000 5.2200

Composition

Component

0.1926 0.0075 Cumene (C9H12) 0.0276 Acetone (C3H6O) 0.7002 α– methylstyrene (C9H10) 0.0721 Total


Composition

13.8046

0.7667

0.5460

0.0303

3.6540

0.2029

18.0046

1.0000

Bottom

Total

72.4100

Acetonephenone (C8H8O) Cumene (C9H12) α– methylstyrene (C9H10) Phenol (C6H5OH) 1.0000 Total

2.0000

0.0368

0.1394

0.0026

1.5660

0.0288

50.7000

0.9319

54.4054

1.0000


53

6.5.3 Separator 5 – Distillation Unit 3 The third distillation unit separates mainly cumene from the other components. Inlet stream from the top product of Distillation Unit 2. The bottom stream is sent to the pre – purification reactor.

Assumptions 1) Distillate composition: 99% of cumene from inlet stream, 20% of α– methylstyrene from inlet stream. 2) Bottom composition: Remaining.


Cumene (C9H12) Acetone (C3H6O) α– methylstyrene (C9H10)

Total

Input Molar Flow Rate (kmol/hr) 13.8046 0.5460 3.6540

18.0046

Composition

Component

0.7667 0.0303 Cumene (C9H12) 0.2029 Acetone (C3H6O) α– methylstyrene (C9H10) Total

Cumene (C9H12) α– methylstyrene (C9H10) 1.0000 Total


Composition

13.6666

0.7649

0.5460

0.0306

3.6540

0.2045

17.8666 Bottom

1.0000

0.1380

0.0451

2.9232

0.9549

3.0612

1.0000


54

6.5.4 Separator 6 – Distillation Unit 4 The fourth distillation unit separates mainly cumene from the other components. Inlet stream from the top product of Distillation Unit 3. The bottom stream is sent to the pre – purification reactor. The top product is sent to acetone product vessel.

Assumptions 1) Distillate composition: 100% of Acetone, 1% of α – methylstyrene and 1% of cumene from inlet stream. 2) Bottom composition: Remaining.


Cumene (C9H12) Acetone (C3H6O) α– methylstyrene (C9H10)

Total

Input Molar Flow Rate (kmol/hr) 13.6666 0.5460 3.6540

17.8666

Composition

Component

0.7649 0.0306 α– methylstyrene (C9H10) 0.2045 Acetone (C3H6O) Cumene (C9H12) Total

Cumene (C9H12) α– methylstyrene (C9H10) 1.0000 Total


Composition

0.0365

0.0508

0.5460

0.7592

0.1367

0.1901

5.3580 Bottom

1.0000

13.5299

0.7890

3.6175

0.2110

0.5070

1.0000


55

6.5.5 Separator 7 – Distillation Unit 5 The fifth distillation unit separates mainly phenol from the other components. Inlet stream from the bottom product of Distillation Unit 2. The bottom stream is sent to the product vessel.

Assumptions 1) Distillate composition: 100% of cumene from inlet stream, 0.01% of phenol from inlet stream, 99% of α– methylstyrene from inlet stream. 2) Bottom composition: Remaining.


Acetonephenone (C8H8O) Cumene (C9H12) α– methylstyrene (C9H10) Phenol (C6H5OH)

Total

Input Molar Flow Rate (kmol/hr) 2.0000 0.1394 1.5660 50.7000

54.4054

Composition

Component

0.0368 0.0026 Cumene (C9H12) 0.0288 α– methylstyrene (C9H10) 0.9319 Phenol (C6H5OH) Total

Acetonephenone (C8H8O) α– methylstyrene (C9H10) Phenol (C6H5OH) 1.0000 Total


Composition

0.1394

0.0823

1.5503

0.9147

0.0051

0.0030

1.6948 Bottom

1.0000

2.0000

0.0379

0.0157

0.0003

50.6949

0.9618

52.7106

1.0000


6.6

56

Reactor 3 – Pre - Purification Reactor

The alpha methyl styrene is converted to cumene by feeding the reactor with hydrogen gas and the aid of Nickel catalyst. This step converts the unsaturated hydrocarbon to saturated hydrocarbon. The unsaturated compound might yield to other than cumene peroxide, this is the reason for this pre – purification method as to eliminate the availability of the unsaturated in the feed stock. The product from this reactor is mix to the fresh cumene feed by a recycle stream. Hydrogen gas is fed with 4:1 ratio with α – methylstyrene fed to the reactor.

Bottom From DU3

H2

Bottom From DU4 n31 + n27 C6 H5 CH(CH3 )2 n32 + n28 C6 H5 C(CH2 )CH3

Pre – Purification Reactor (R3)

n37 C6 H5 CH(CH3 )2

Chemical Reaction Involved In Pre – Purification Reactor 1) C6 H5 C(CH2 )CH3 (ℓ) + H2 (g) → C6 H5 CH(CH2 )CH3 (ℓ)__________________ξ4 α – methylstyrene + Hydrogen gas → Cumene

Material Balance Type of reaction Component

Cumene (C9H12) α – methylstyrene (C9H10) Hydrogen (H2) Total

= Reactive reaction Input Molar Composition Component Flow Rate (kmol/hr) 13.6679 0.2947 Cumene (C9H12) 6.5407 0.1411 26.1628

0.5642

46.3714

1.0000

Total

Output Molar Composition Flow Rate (kmol/hr) 20.2086 1.0000

20.2086

1.0000


7.0

ENERGY BALANCE

7.1

Sample Calculation

Assumption in calculation 1) The open system at steady state 2) No moving parts in the system 3) The linear velocities of all streams at single height. 4) The effects of pressure will be neglect if the pressure of the system under 30 atm. 5) The heat of formation for certain substances will be neglected since the limitation of physical properties data. 6) All the physical and chemical properties of substances was refer from Perry’s Chemical Engineer’s Handbook, Perry, R.H. and Elementary Chemical Process.

General Energy Balance Equation Q − Wt = ∆H + ∆Ek + ∆Ep

Based on the assumptions stated above, Wt = 0,

∆Ek = 0,

∆Ep = 0

Hence, the equation can be reduced to Q = ∆H

Equation for reactive system: ̌ 𝐭 + ∆𝐇 ̌ 𝐟 ) − ∑ 𝐧𝐢 (∆𝐇 ̌ 𝐭 + ∆𝐇 ̌ 𝐟) ∆𝐇 = ∑ 𝐧𝐢 (∆𝐇 𝐨𝐮𝐭𝐥𝐞𝐭

𝐢𝐧𝐥𝐞𝐭

Equation for non-reactive system: ̌ 𝐭 ) − ∑ 𝐧𝐢 (∆𝐇 ̌ 𝐭) ∆𝐇 = ∑ 𝐧𝐢 (∆𝐇 𝐨𝐮𝐭𝐥𝐞𝐭

𝐢𝐧𝐥𝐞𝐭

57

PLANT DESIGN FOR PRODUCTION OF PHENOL T

̌t = ∫ Which, ∆H T

ref

Cp dT T

∫T

ref

T

Cp dT = ∫T (a + bT + cT 2 + dT 3 + eT 4 )dT ref

Where, a, b, c, d, e = the heat capacities from Perry’s handbooks The reference conditions for the system: i.

Tref = 25°C

ii.

Pref = 1 atm

7.1.1 Pump 1 – Cumene Feed Pump ∆𝐻2 T = 25°C P = 6 atm

∆𝐻1 T = 25°C P = 1 atm

Energy balance for Pump 1: ̌ t ) − ∑ ni (∆H ̌ t) ∆H = ∑ ni (∆H outlet

inlet

Reference state: C9H12 (l, 25°C, 1 atm) Substances

Cumene(C6 H5 CH(CH3 )2 )

Inlet

Outlet

nin (kmol/h)

̌ in (kJ/h) H

nout (kmol/h)

̌ out (kJ/h) H

100

H1

100

H2

Inlet: ̌ t ) = 𝟎 𝐤𝐉 (Since the stream in references condition) ∆H1 = ∑inlet ni (∆H 𝐡 Outlet: ̌ t ) = 𝟎 𝐤𝐉 (Heat changes from pressure differences was neglected) ∆H2 = ∑outlet ni (∆H 𝐡 ̌ t ) − ∑inlet ni (∆H ̌ t ) = 0 kW (Power for Pump 1) ∆H = ∑outlet ni (∆H

58

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.2 Compressor 1 – Air Feed

Energy balance for Compressor 1: ̌𝑡 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡

𝑖𝑛𝑙𝑒𝑡

Reference state: O2, N2 (g, 25°C, 1 atm) Substances

Inlet

Outlet

nin (kmol/h)

̌ in (kJ/h) H

nout (kmol/h)

̌ out (kJ/h) H

Oxygen (O2)

400

H3a

400

H4a

Nitrogen (N2)

1504.76

H3b

1504.76

H4b

Inlet: ̌ t ) = 𝟎 𝒌𝑱 (Since the stream in references condition) ∆H3 = ∑inlet ni (∆H 𝒉 Outlet: ̌ t ) = 𝟎 𝐤𝐉 (Heat changes from pressure differences was neglected) ∆H4 = ∑outlet ni (∆H 𝐡 ̌ t ) − ∑inlet ni (∆H ̌ t ) = 0 kW (Power for Compressor 1) ∆H = ∑outlet ni (∆H

59

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.3 Reactor 1 – Oxidation Reactor

∆𝐻5 T = 100°C, P = 6 atm

NaOH

∆𝐻6

∆𝐻2 Oxidation Reactor (R1)

∆𝐻4

T = 100°C, P = 6 atm

Energy balance for Reactor 1: ̌𝑡 + ∆𝐻 ̌𝑓 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 + ∆𝐻 ̌𝑓 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡


Reference state: O2, N2, H2 (g, 25°C, 1 atm) and C(s, 25°C, 1 atm) Substances

Inlet nin (kmol/h)

Outlet ̌ in H

nout (kmol/h)

(kJ/h)

̌ out H (kJ/h)


100

H2

14

H6a

Oxygen (O2 )

400

H4a

317

H5a

Nitrogen (N2 )

1504.76

H4b

1504.76

H5b

Methanol (CH3 OH)

-

-

2

H5c

Cumene Peroxide( C6 H5 COOH(CH3 )2)

-

-

78

H6b

DMBA(C6 H5 COH(CH3 )2 )

-

-

6

H6c

Acetonephenone (C6 H5 COCH3 )

-

-

2

H6d

Inlet: 𝐤𝐉⁄ 𝐡 𝐤𝐉 ∆H4 = 𝟎 ⁄𝐡 ∆H2 = 𝟎

Outlet: ∆H5 = nH5a + nH5b + nH5c 100

nH5a = 𝑛𝑖 (∫ 25

̌𝑓 ) = 317 Cp dT + ∆𝐻

𝑘𝑚𝑜𝑙 kJ 𝑚𝑜𝑙 𝑘𝑗 2.24 + 0 (1000 ) = 710080 ℎ mol 𝑘𝑚𝑜𝑙 ℎ

60

PLANT DESIGN FOR PRODUCTION OF PHENOL 100

̌𝑓 ) = 1504.76 Cp dT + ∆𝐻

nH5b = 𝑛𝑖 (∫ 25

= 3295424.4

𝑘𝑗 ℎ

100

̌𝑓 ) = 2 Cp dT + ∆𝐻

nH5c = 𝑛𝑖 (∫ 25

= −395200

𝑘𝑚𝑜𝑙 kJ 𝑚𝑜𝑙 2.19 + 0 (1000 ) ℎ mol 𝑘𝑚𝑜𝑙

𝑘𝑚𝑜𝑙 𝑘𝑗 𝑘𝐽 𝑚𝑜𝑙 (3.6025 − 201.2 ) 1000 ℎ 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑘𝑚𝑜𝑙

𝑘𝑗 ℎ

∆H5 = nH5a + nH5b + nH5c = 710080 + 3295424.4 − 395200 = 𝟑𝟔𝟏𝟎𝟑𝟎𝟒. 𝟒

𝒌𝒋 𝒉

∆H6 = nH6a + nH6b + nH6c + nH6d 100

̌𝑓 ) = 14 Cp dT + ∆𝐻

nH6a = 𝑛𝑖 (∫ 25

= 40351.08

̌𝑓 ) = 78 Cp dT + ∆𝐻

25

= 332560800

̌𝑓 ) = 6 Cp dT + ∆𝐻

25

= 21358800

𝑘𝑚𝑜𝑙 kJ 4263.6 −0 ℎ mol

1000

𝑚𝑜𝑙 𝑘𝑚𝑜𝑙


1000



1000

𝑚𝑜𝑙 𝑘𝑗 = 4929000 𝑘𝑚𝑜𝑙 ℎ

𝑘𝑗 ℎ

100

̌𝑓 ) = 2 Cp dT + ∆𝐻

nH6d = 𝑛𝑖 (∫


𝑘𝑗 ℎ

100

nH6c = 𝑛𝑖 (∫

1000

𝑘𝑗 ℎ

100

nH6b = 𝑛𝑖 (∫

𝑘𝑚𝑜𝑙 𝑘𝐽 𝑘𝐽 2920.62 − 38.40 ℎ 𝑚𝑜𝑙 𝑚𝑜𝑙

25

∆H6 = nH6a + nH6b + nH6c + nH6d = 40351.08 + 332560800 + 21358800 + 4929000 = 𝟑𝟓𝟖𝟖𝟖𝟖𝟗𝟓𝟏. 𝟏

∆H = (∆H5 + ∆H6 ) − (∆H2 + ∆H4 ) = 358888951.1

𝑘𝐽 ℎ ℎ 3600𝑠

= 99691.37 𝑘𝑊 (𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 1)

𝒌𝑱 𝒉

61

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.4 Reactor 2 – Cleavage Unit H2 SO4

∆𝐻6

∆𝐻7 T = 110°C, P = 6 atm

Cleavage Unit (R2)

Energy balance for Reactor 2: ̌𝑡 + ∆𝐻 ̌𝑓 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 + ∆𝐻 ̌𝑓 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡


Reference state: O2, N2, H2 (g, 25°C, 1 atm) and C(s, 25°C, 1 atm) Substances

Inlet

Outlet

nin

̌ in H

(kmol/h)

(kJ/h)


14

H6a

14

H7a

Cumene Peroxide( C6 H5 COOH(CH3 )2)

78

H6b

-

-

DMBA(C6 H5 COH(CH3 )2 )

6

H6c

-

-


2

H6d

2

H7b

Phenol (C6 H5 OH)

-

-

50.7

H7c

Acetone ((CH3 )2 CO)

-

-

27.3

H7d

α − methylstyrene(C6 H5 C(CH2 )CH3 )

-

-

5.22

H7e

Water (H2 O)

-

-

0.78

H7f

nout (kmol/h)

(kJ/h)

Inlet: kJ

∆H6 = 358888951.1 h Outlet: ∆H7 = 151857055.6

kJ h

∆H = ∆H7 − ∆H6 = −207031895.5

̌ out H


= −57508.86 𝑘𝑊 ( 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑜𝑟 2)

62

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.5 Heat Exchanger 1 – Cooling

∆𝐻7

∆𝐻8 T = 25°C, P = 6 atm

Heat exchanger 1

Energy balance for Heat exchanger 1: ̌𝑡 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡


Reference state: liquid substances at 25°C, 1 atm Substances

Inlet nin (kmol/h)

Outlet ̌ in H

nout (kmol/h)

(kJ/h)

̌ out H (kJ/h)


14

H7a

14

H8a


2

H7b

2

H8b

Phenol (C6 H5 OH)

50.7

H7c

50.7

H8c


27.3

H7d

27.3

H8d


5.22

H7e

5.22

H8e

Water (H2 O)

0.78

H7f

0.78

H8f

Inlet: ∆H7 = 151857055.6

kJ h

Outlet: ∆H8 = 0

kJ h

∆H = ∆H8 − ∆H7 = 151857055.6

kJ ℎ h 3600𝑠

= 42182.52 𝑘𝑊 ( 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐻𝑒𝑎𝑡 𝐸𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 1)

63

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.6 Separator 1 - Phase Separator (Gravity Settler)

∆𝐻9

T = 25°C, P = 6 atm

∆𝐻8

Phase Separator (S1)

H2 SO4 Energy balance for Phase Separator (S1): ̌𝑡 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡



Inlet

Outlet


nin (kmol/h) 14

̌ in H (kJ/h) H8a

nout (kmol/h) 14

̌ out H (kJ/h) H9a


2

H8b

2

H9b

Phenol (C6 H5 OH)

50.7

H8c

50.7

H9c


27.3

H8d

27.3

H9d


5.22

H8e

5.22

H9e

Water (H2 O)

0.78

H8f

0.78

H9f

Inlet: ∆H8 = 0

kJ h

Outlet: ∆H9 = 0

kJ h

∆H = ∆H9 − ∆H8 = 0 𝑘𝑊 ( 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑃ℎ𝑎𝑠𝑒 𝑆𝑒𝑝𝑎𝑟𝑎𝑡𝑜𝑟)

64

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.7 Separator 2 – Scrubber H2 O ∆𝐻10

∆𝐻9

T = 100°C, P = 1 atm

Scrubber (S2)

Acidified Wash Water Energy balance for Scrubber (S2): ̌𝑡 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡



Inlet

Outlet

nin

̌ in H

(kmol/h)

(kJ/h)


14

H9a

14

H10a


2

H9b

2

H10b

Phenol (C6 H5 OH)

50.7

H9c

50.7

H10c


27.3

H9d

27.3

H10d


5.22

H9e

5.22

H10e

Water (H2 O)

0.78

H9f

0.78

H10f

nout (kmol/h)

̌ out H (kJ/h)

Inlet: ∆H9 = 0

kJ h

Outlet: ∆H10 = 136337110

kJ h

∆H = ∆H10 − ∆H9 = 136337110

kJ ℎ = 37871.42 𝑘𝑊 ( 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑆𝑐𝑟𝑢𝑏𝑏𝑒𝑟) h 3600𝑠

65

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.8 Heat Exchanger 2 – Cooling ∆𝐻10

∆𝐻11 T = 80°C, P = 1 atm

Heat exchanger 2

Energy balance for Heat exchanger 2: ̌𝑡 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡



Inlet

Outlet

nin

̌ in H

(kmol/h)

(kJ/h)


14

H10a

14

H11a


2

H10b

2

H11b

Phenol (C6 H5 OH)

50.7

H10c

50.7

H11c


27.3

H10d

27.3

H11d


5.22

H10e

5.22

H11e

Water (H2 O)

0.78

H10f

0.78

H11f

Inlet: ∆H10 = 136337110

kJ h

Outlet: ∆H11 = 99980603.9

kJ h

∆H = ∆H11 − ∆H10 = −36356506.1


= −10099.03 𝑘𝑊 ( 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐻𝑒𝑎𝑡 𝐸𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 2)

nout (kmol/h)

̌ out H (kJ/h)

66

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.9 Separator 3 – Distillation Unit 1 ∆𝐻12

Distillation Unit 1

T = 100°C, P = 1 atm

∆𝐻11

∆𝐻13 T = 100°C, P = 1 atm

Energy balance for distillation unit 1: ̌𝑡 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡




Inlet

Outlet

nin

̌ in H

(kmol/h)

(kJ/h)

14

H11a


nout (kmol/h)

̌ out H (kJ/h)

0.056

H12a

13.94

H13a


2

H11b

2

H13b

Phenol (C6 H5 OH)

50.7

H11c

50.7

H13c


27.3

H11d

26.75

H12b

0.55

H13d

Acetone ((CH3 )2 CO) α − methylstyrene(C6 H5 C(CH2 )CH3 )

5.22

H11e

5.22

H13e

Water (H2 O)

0.78

H11f

0.78

H12c

67

PLANT DESIGN FOR PRODUCTION OF PHENOL Inlet: ∆H11 = 99980603.9

kJ h

Outlet: ∆H12 = 415108.87

kJ h

∆H13 = 135910318.3

kJ h

∆H = (∆H12 + ∆H13 ) − ∆H11 = 36344823.22


= 10095.78 𝑘𝑊 ( 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐷𝑖𝑠𝑡𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝑈𝑛𝑖𝑡 1)

7.1.10 Separator 4 – Distillation Unit 2 ∆𝐻14

Distillation Unit 2

T = 175°C, P = 1 atm

∆𝐻13

∆𝐻15 T = 175°C, P = 1 atm



68

PLANT DESIGN FOR PRODUCTION OF PHENOL Reference state: liquid substances at 25°C, 1 atm Substances


Inlet

Outlet

nin

̌ in H

(kmol/h)

(kJ/h)

13.94

H13a


nout (kmol/h)

̌ out H (kJ/h)

13.8

H14a

0.14

H15a


2

H13b

2

H15b

Phenol (C6 H5 OH)

50.7

H13c

50.7

H15c


0.55

H13d

0.55

H14b


5.22

H13e

3.65

H14c

1.57

H15d

α − methylstyrene(C6 H5 C(CH2 )CH3 ) Inlet: ∆H13 = 135910318.3

kJ h

Outlet: ∆H14 = 99225378

kJ h

∆H15 = 172595120.2

kJ h

∆H = (∆H14 + ∆H15 ) − ∆H13 = 135910179.9

kJ h h 3600s

= 37752.83 kW ( Power of Distillation Unit 2)

69

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.11 Heat Exchanger 3 – Cooling ∆𝐻16

∆𝐻14

T = 150°C, P = 1 atm

Heat exchanger 3

Energy balance for heat exchanger 3: ̌𝑡 ) − ∑ 𝑛𝑖 (∆𝐻 ̌𝑡 ) ∆H = ∑ 𝑛𝑖 (∆𝐻 𝑜𝑢𝑡𝑙𝑒𝑡



Inlet

Outlet

nin

̌ in H

(kmol/h)

(kJ/h)


13.80

H14a

13.80

H16a


0.55

H14b

0.55

H16b


3.65

H14c

3.65

H16c

Inlet: ∆H14 = 99225378

kJ h

Outlet: ∆H16 = 82689030

kJ h

∆H = ∆H16 − ∆H14 = −16536348

kJ h h 3600s

= −4593.43 kW ( Power of Heat Exchanger 3)

nout (kmol/h)

̌ out H (kJ/h)

70


Distillation Unit 3

T = 152°C, P = 1 atm

∆𝐻16

∆𝐻18 T = 152°C, P = 1 atm




𝐶𝑢𝑚𝑒𝑛𝑒(C6 H5 CH(CH3 )2 )

Inlet

Outlet

nin

̌ in H

(kmol/h)

(kJ/h)

13.80

H16a


nout (kmol/h)

̌ out H (kJ/h)

13.67

H17a

0.13

H18a

𝐴𝑐𝑒𝑡𝑜𝑛𝑒 ((CH3 )2 CO)

0.55

H16b

0.55

H17b

𝛼 − 𝑚𝑒𝑡ℎ𝑦𝑙𝑠𝑡𝑦𝑟𝑒𝑛𝑒(C6 H5 C(CH2 )CH3 )

3.65

H16c

3.65

H17c

2.92

H18b

𝛼 − 𝑚𝑒𝑡ℎ𝑦𝑙𝑠𝑡𝑦𝑟𝑒𝑛𝑒(C6 H5 C(CH2 )CH3 ) Inlet: ∆H16 = 82689030

kJ h

Outlet: kJ h kJ = 13245557.8 h

∆H17 = 83369109.6 ∆H18

∆H = (∆H17 + ∆H18 ) − ∆H16 = 13925637.4

kJ h h 3600s


71


Distillation Unit 4

T = 100°C, P = 1 atm

∆𝐻17

∆𝐻20 T = 100°C, P = 1 atm





Inlet

Outlet

nin

̌ in H

nout (kmo

̌ out H

(kmol/h)

(kJ/h)

l/h)

(kJ/h)

13.67

H16a

0.14

H18a

13.53

H19a

𝐶𝑢𝑚𝑒𝑛𝑒(C6 H5 CH(CH3 )2 ) 𝐴𝑐𝑒𝑡𝑜𝑛𝑒 ((CH3 )2 CO)

0.55

H16b

0.55

H18b

𝛼 − 𝑚𝑒𝑡ℎ𝑦𝑙𝑠𝑡𝑦𝑟𝑒𝑛𝑒(C6 H5 C(CH2 )CH3 )

3.65

H16c

0.037

H18c

3.62

H19b

𝛼 − 𝑚𝑒𝑡ℎ𝑦𝑙𝑠𝑡𝑦𝑟𝑒𝑛𝑒(C6 H5 C(CH2 )CH3 ) Inlet: ∆H17 = 83369109.6

kJ h

Outlet: kJ h kJ = 48742644.6 h

∆H19 = 508273.85 ∆H20

∆H = (∆H19 + ∆H20 ) − ∆H17 = −34118191.15


= −9477.28 𝑘𝑊 ( 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐷𝑖𝑠𝑡𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝑈𝑛𝑖𝑡 4)

72

PLANT DESIGN FOR PRODUCTION OF PHENOL 7.1.14 Separator 7 – Distillation Unit 5

∆𝐻21

Distillation Unit 5

T = 180°C, P = 1 atm

∆𝐻15

∆𝐻22 T = 180°C, P = 1 atm




Inlet

Outlet

nin (kmol/h)

̌ in H (kJ/h)

nout (kmol/h)

̌ out H (kJ/h)


0.14

H15a

0.14

H21a


2

H15b

2

H22a

Phenol (C6 H5 OH)

50.7

H15c

0.005

H21b

50.695

H22b

1.55

H21c

0.02

H22c

Phenol (C6 H5 OH) α − methylstyrene(C6 H5 C(CH2 )CH3 )

1.57

H15d

α − methylstyrene(C6 H5 C(CH2 )CH3 ) Inlet: ∆H15 = 172595120.2

kJ h

Outlet: kJ h kJ = 169324023.7 h

∆H21 = 9024639.35 ∆H22

∆H = (∆H21 + ∆H22 ) − ∆H15 = 5753542.8

kJ h h 3600s


73


8.0

SIZING OF EQUIPMENTS

8.1

Sizing Of The Equipments

The flow sheet of cumene oxidation is shown in process flow diagram. The oxidation of cumene as the first reaction usually carried out in a bubble column reactor with temperature inlet is 25°C and pressure 6 atm meanwhile, the product will be yielded at temperature 100°C and pressure remain the same. Fresh cumene are fed to the first reactor (R-101). Oxygen is used in each reactor as oxidant. For first reactor, the bubble column reactor is chosen as the relevant equipment. This is because we have oxidation process occur at first reactor. Based on (Mengmeng & Ekaterina, 2017) they stated in their article, bubble column reactor is used in chemical process which involved oxidation, chlorination, polymerization and hydrogenation. This bubble column reactor also was chosen because of their technical structure. Founded by (Kantarci, Borak, & Ulgen, 2005) in (Mengmeng & Ekaterina, 2017) article, bubble column reactor has high heat and mass transfer coefficient, need little maintenance and have low operating cost due to lack of moving parts and compactness compared to batch reactor. Before choosing the bubble column reactor, the three main phenomena which are 1) heat and mass transfer characteristics 2) mixing characteristic and 3) system have been taking in considered before establishing the decision to use this type of reactor. For reactor R-102 and R-103, the reactor use is Continous Stirred Tank Reactor (CSTR). Beside, can produce the large amount of our product which is Phenol, according to (Chaplin, 2014), CSTR is an easily constructed, versatile and cheap reactor. It allows simple catalyst (Nickel, Ni in R-103) charging and replacement. It is well-mixed nature permits straightforward control over the temperature and pH of the reaction and able to supply or remove unwanted gasses easily. He continued that, CSTR tend to be as large as need to be efficiently mixed to yield product. CSTR has the advantage that there is very little resistance to the flow of the reactant stream which might contain colloidal or insoluble reactants. Furthermore, the mechanical of its structure which is stirring limits which will support the immobilized enzymes to easily disintegrate to give ‘fines’ shape to enter the product stream. In addition, CSTR has complete back-mixing which will results in minimization of the reactant concentration but at the same time, maximize the product concentration, relative to final conversion, at every point within the effectiveness factor being uniform throughout. Thus, CSTR are the preferred reactors, everything being equal in the reactor for process involving reactant inhibition or product activation (Chaplin, 2014).

74

PLANT DESIGN FOR PRODUCTION OF PHENOL 8.1.1 Reactor 1 – Oxidation Reactor CSTR Volumetric Flowrate, vo =

Mass Flowrate ( Density (

kg ) h

kg ) m3

V = space time, τ × vo 1. vo cumene = 14

kmol hr

kg

kg

× 120.19 kmol ÷ 862 m3

m3 = 1.952 hr

2. vo cumene peroxide = 78

kmol

= 11470.59 3. vo DMBA = 6

kmol hr

× 150

hr m3

kg kmol

÷ 1.02

kg m3

hr kg

÷ 1013 m3 × 256.35

kg kmol

m3 = 1.518 hr 4. vo Acetonephenone = 2

kmol hr

÷ 1.03

kg m3

× 120.15

kg kmol

m3 = 233.01 hr 5. Total volumetric, 𝐯𝐨 = 11704.51

𝐦𝟑 𝐡𝐫

From Rule 5 Heuristics for Reactor, table 11.17; Ideal CSTR behavior is approached when the mean of residence time is 5 to 10 times the length needed to achieve homogeneity, which is accomplished with 500-2000 revolutions of a properly designed stirrer. We assumed that,𝛕 = 𝟎. 𝟎𝟏𝟓 𝐦𝐢𝐧, 𝐥 = 𝟎. 𝟏𝟓 𝐦; 𝐕 = 𝟐. 𝟗𝟑 𝐦𝟑 V = πr 2 l 2.93 = π r 2 (0.15) 𝐫 = 𝟐. 𝟒𝟗 𝐦

75

PLANT DESIGN FOR PRODUCTION OF PHENOL 8.1.2 Reactor 2 – Cleavage Unit CSTR kmol

1. vo cumene = 14

kg

kg

× 120.19 kmol ÷ 862 m3

hr

m3 = 1.952 hr 2. vo Acetonephenone = 2

kmol hr

÷ 1.03

kg m3

× 120.15

kg kmol

m3 = 233.01 hr 3. vo phenol = 50.7

kmol

= 4745.43 4. vo acetone = 27.3

hr 3

kg

kg

÷ 1.07 m3 × 100.15 kmol

m hr

kmol hr

÷ 784

kg m3

kg

× 58.08

kmol

m3 = 2.022 hr 5. vo α − methylstyrene = 5.22

kmol hr 3

÷ 910

m hr kmol kg 6. vo water = 0.78 hr ÷ 997 m3 × 18.02

kg m3

× 118.18

kg kmol

= 0.678

kg kmol

m3 = 0.0141 hr 7. Total volumetric, 𝐯𝐨 = 4983.11

𝐦𝟑 𝐡𝐫

From Rule 5 Heuristics for Reactor, table 11.17; Ideal CSTR behavior is approached when the mean of residence time is 5 to 10 times the length needed to achieve homogeneity, which is accomplished with 500-2000 revolutions of a properly designed stirrer. We assumed that,𝛕 = 𝟎. 𝟑𝟓 𝐦𝐢𝐧, 𝐥 = 𝟑. 𝟓 𝐦; 𝐕 = 𝟐𝟗. 𝟎𝟔𝟖 𝐦𝟑 V = πr 2 l 29.068 = π r 2 (3.5) 𝐫 = 𝟏. 𝟔𝟑 𝐦

76

PLANT DESIGN FOR PRODUCTION OF PHENOL 8.1.3 Reactor 3 – Pre-Purification Reactor CSTR kmol

1. vo cumene = 14 = 𝟐. 𝟔𝟕

hr 𝟑

kg

kg

× 120.19 kmol ÷ 862 m3

𝐦 𝐡𝐫

Ideal CSTR behavior is approached when the mean of residence time is 5 to 10 times the length needed to achieve homogeneity, which is accomplished with 500-2000 revolutions of a properly designed stirrer. We assumed that,𝛕 = 𝟐. 𝟐𝟓 𝐦𝐢𝐧, 𝐥 = 𝟓 𝐦; 𝐕 = 𝟎. 𝟏 𝐦𝟑 V = πr 2 l 0.1 = π r 2 (5) 𝐫 = 𝟎. 𝟎𝟖 𝐦

8.1.4 Separator 1 - Phase Separator (Gravity Settler) Density (ρ), kg/m3

Cumene (C9H12)

Molecular weight (MW), kg/kmol 120.19

Acetophenone (C8H8O)

120.15

1.03

Phenol (C6H5OH)

100.15

1.07

Acetone (C3H6O)

58.08

784.00

α– methylstyrene (C9H10)

118.18

910.00

Water (H2O)

18.02

997.00

Sulphuric Acid (H2SO4)

98.08

1.84

Substances

862.00

Temperature inlet

25.0 °C

Pressure inlet

6.0 atm

Temperature outlet

25.0 °C

Pressure outlet

6.0 atm

77


From Table 11.6, the following heuristics are used: Rule 3 – Vertical vessel. Rule 4 – L/D between 2.5 and 5 with optimum at 3.0. Rule 5 – Liquid holdup time is 5 min based on ½ volume of vessel. Rule 9 – Gas velocity, u is given by

u = k√

ρv − 1 m/s ρl

Where k = 0.0305(0.1) for vessels without mesh entrainer. Rule 12 – Good performance obtained at 30% - 100% of u from Rule 9; typical value is 75%.

From Table 1.5, Vapour flow = 9,217.11 kg/h, P = 6.01325 bar, T = 25°C Liquid flow = 98.08 kg/h, P = 6.01325 bar, T = 25°C ρv = 335.28

kg , m3

ρl = 1.84

kg m3

From Rule 9, 335.28 m u = 0.00305√ − 1 m/s = 0.0411 1.84 s Use uact = (0.75)(0.0411 m/s) = 0.0308 m/s Thus, uρv πD2 = mass flowrate of vapour 4

78


kg m (0.0308 s ) 335.28 3 πD2 kg m = 9217.11 4 h 103.27

1h 3600 s

kg πD2 kg m2 s = 2.56 4 s

kg 2.56 s (4) D=√ = √0.3157 m2 = 0.56 m kg 103.27 2 π m s

From Rule 5,

Volume of liquid =

0.5 LπD2 0.5 Lπ(0.56)2 = = 0.1232L m2 4 4

From Rule 4, L/D should be in the range 2.5 to 5. For this case, L/D = 2.29 m3 kg 60 s 1h 1 min) 9217.11 h (3600 s) = 2.29 m3 kg 335.28 3 m

(5 min) ( 5 min of vapour flow =

Equation the two results above, 0.1232L m2 = 2.29 m3 2.29 m3 L= = 18.59 m 0.1232 m2 From Rule 4, L/D should be in the range 2.5 to 5. For this case, L/D = 18.59/0.56 = 33. Because this is out of range, change to L = 2.5D = 2.5(0.56 m) = 1.4 m. Heuristics from Table 11.6 suggest that Phase Separator should be a vertical vessel with D = 0.56 m and L = 1.4 m. Thus, the L/D is to be in the minimum range which is 2.5. It can be concluded that the design of Phase Separator is consistent with the heuristics given in Table 11.6. The small differences in L and D are to be expected in a comparison such as this calculation.

79

PLANT DESIGN FOR PRODUCTION OF PHENOL 8.1.5 Separator 2 - Scrubber Scrubber is used to wash the acid with water. The water extracts the remaining acids in the organic stream and leaves as acidified wash water. Density (ρ), kg/m3

Cumene (C9H12)

Molecular weight (MW), kg/kmol 120.19

Acetophenone (C8H8O)

120.15

1.03

Phenol (C6H5OH)

100.15

1.07

Acetone (C3H6O)

58.08

784.00

α– methylstyrene (C9H10)

118.18

910.00

Water (H2O)

18.02

997.00

Substances

862.00

Temperature inlet

25.0 °C

Pressure inlet

6.0 atm

Temperature outlet

100.0 °C

Pressure outlet

1.0 atm

From Table 11.6, the following heuristics are used: Rule 3 – Vertical vessel. Rule 4 – L/D between 2.5 and 5 with optimum at 3.0. Rule 5 – Liquid holdup time is 5 min based on ½ volume of vessel. Rule 9 – Gas velocity, u is given by

u = k√

ρl − 1 m/s ρv

Where k = 0.0305(0.1) for vessels without mesh entrainer. Rule 12 – Good performance obtained at 30% - 100% of u from Rule 9; typical value is 75%.

80

PLANT DESIGN FOR PRODUCTION OF PHENOL From Table 1.5, Vapour flow = 9,217.11 kg/h, P = 6.01325 bar, T = 25°C Liquid flow = 98.08 kg/h, P = 6.01325 bar, T = 25°C ρv = 335.28

kg , m3

ρl = 1.84

kg m3

From Rule 9, 335.28 m u = 0.00305√ − 1 m/s = 0.0411 1.84 s Use uact = (0.75)(0.0411 m/s) = 0.0308 m/s Thus, uρv πD2 = mass flowrate of vapour 4 kg m (0.0308 s ) 335.28 3 πD2 kg m = 9217.11 4 h 103.27

1h 3600 s

kg πD2 kg m2 s = 2.56 4 s

kg 2.56 s (4) D=√ = √0.3157 m2 = 0.56 m kg 103.27 2 π m s

From Rule 5,

Volume of liquid =

0.5 LπD2 0.5 Lπ(0.56)2 = = 0.1232L m2 4 4

81


From Rule 4, L/D should be in the range 2.5 to 5. For this case, L/D = 2.29 m3 kg 60 s 1h ) 9217.11 ( ) 1 min h 3600 s = 2.29 m3 kg 335.28 3 m

(5 min) ( 5 min of vapour flow =

Equation the two results above, 0.1232L m2 = 2.29 m3

L=

2.29 m3 = 18.59 m 0.1232 m2

From Rule 4, L/D should be in the range 2.5 to 5. For this case, L/D = 18.59/0.56 = 33. Because this is out of range, change to L = 2.5D = 2.5(0.56 m) = 1.4 m. Heuristics from Table 11.6 suggest that scrubber should be a vertical vessel with D = 0.56 m and L = 1.4 m. Thus, the L/D is to be in the minimum range which is 2.5. It can be concluded that the design of scrubber is consistent with the heuristics given in Table 11.6. The small differences in L and D, are to be expected in a comparison such as this calculation.

8.1.6 Separator 3 – Distillation Column Unit 1 Distillation column 1 is used for separating acetone. Thus, it is also known as Acetone column. Rule 2 (Table 11.13) : Relative volatility αavg =

P1sat P2sat

For multi-component system, the relative volatility is as follows: αavg = 3√αFij αBij αDij In order to find the vapor pressure of each component, Antoine equation is used as below: log10 P (mmHg) = A −

B T(˚C) + C

82

PLANT DESIGN FOR PRODUCTION OF PHENOL For Feed: Vapor pressure of Acetone at 80˚C, P1sat : log10 P1sat = 7.2316 −

1277.03 80 + 237.23

P1sat = 1607.0706 mmHg P1sat = 2.1146 atm Vapor pressure of Cumene at 80˚C, P2sat : log10 P2sat = 6.93666 −

1460.793 80 + 207.777

P2sat = 72.5323 mmHg P2sat = 0.0954 atm Relative volatility for feed: αFij =

P1sat 2.1146 atm = P2sat 0.0954 atm

𝛂𝐅𝐢𝐣 = 𝟐𝟐. 𝟏𝟔𝟓𝟔 For Distillate: Vapor pressure of Acetone at 100˚C, P1sat : log10 P1sat = 7.2316 −

1277.03 100 + 237.23

P1sat = 2784.6970 mmHg P1sat = 3.6641 atm Vapor pressure of Cumene at 100˚C, P2sat : log10 P2sat = 6.93666 −

1460.793 100 + 207.777

P2sat = 155.0205mmHg P2sat = 0.2040 atm

83

PLANT DESIGN FOR PRODUCTION OF PHENOL Relative volatility for feed: αDij

P1sat 3.6641 atm = sat = 0.2040 atm P2 𝛂𝐃𝐢𝐣 = 𝟏𝟕. 𝟗𝟔𝟏𝟑

For Bottom: The relative volatility of the bottom part of the column is same as the distillate because its temperature condition is same which is 100˚C. Thus, 𝛂𝐁𝐢𝐣 = 𝛂𝐃𝐢𝐣 = 𝟏𝟕. 𝟗𝟔𝟏𝟑 The average volatility is: 3

αavg = √22.1656(17.9613)(17.9613) 𝛂𝐚𝐯𝐠 = 𝟏𝟗. 𝟐𝟔𝟓𝟕 Rule 7 (Table 11.13) : The minimum number of trays with Fenske-Underwood equation. For multi-component system using Fenske equation: ln [ Nmin = ln [ Nmin =

(xDi /xBi ) ] (xDj /xBj −1 ln α

(0.9697/0.0075) ] (0.0020/0.1926 −1 ln 19.2657

𝐍𝐦𝐢𝐧 = 𝟐. 𝟏𝟖𝟕𝟓 Rule 6 (Table 11.13) : The optimum number of theoretical trays is near 2Nmin Nopt = 2Nmin = 4.375 Rule 9 (Table 11.13) : A safety factor of 10% on number of trays Nactual =

2Nmin (1 + 0.1) ε

Nactual =

4.375(1 + 0.1) 0.85

𝐍𝐚𝐜𝐭𝐮𝐚𝐥 = 𝟓. 𝟔𝟔

84

PLANT DESIGN FOR PRODUCTION OF PHENOL Rule 4 (Table 11.14) : Tray efficiencies for distillation of light hydrocarbons and aq. Solution ε = 60 → 90% It is assumed that the tray efficiencies is 𝛆 = 𝟖𝟓% Rule 8 (Table 11.13) : Minimum reflux at bubble point

R min

kmol kmol (100 /27.59 ) (F/D) hr hr = = (α − 1) (19.2657 − 1) 𝐑 𝐦𝐢𝐧 = 𝟎. 𝟏𝟗𝟖𝟒

Rule 5 (Table 11.13) : The optimum reflux in the range of 1.2-1.5R min R = (1.2 → 1.5)R min The optimum reflux is assumed to be 1.35𝐑 𝐦𝐢𝐧 R = (1.35)R min = 1.35(0.1984) 𝐑 = 𝟎. 𝟐𝟔𝟕𝟖 Rule 2 (Table 11.14) : Peak efficiency of vapor factor in range 1.2-1.5

m s

[kg/m3 ]0.5

Fs = uρ0.5 v The efficiency of vapor factor is assumed to be 1.35

𝐦 𝐬

[𝐤𝐠/𝐦𝟑 ]𝟎.𝟓

In order to find the density of the vapor in the distillate, equation below is used: ρn = ∑(xi ρi )n For substance at distillate: Substances

Molar

Molecular

Mass

Mass

Density,

flowrate

weight

flowrate

fraction,

ρ (kg/m3)

(kmol/hr)

(kg/kmol)

(kg/hr)

x

Cumene

0.0560

120.19

Acetone

26.7540

Water

0.7800

Total

xρ

6.7306

0.0043

862.00

3.7066

58.08 1553.8723

0.9868

784.00

773.6512

18.02

14.0556

0.0089

997.00

8.8733

1574.6585

1.0000

786.2311

85

PLANT DESIGN FOR PRODUCTION OF PHENOL ρv = 786.2311 kg/m3 u=

Fs 1.35 = ρ0.5 786.23110.5 v

𝐮 = 𝟎. 𝟎𝟒𝟖𝟏 𝐦/𝐬 The diameter, D of the tower is assumed to be 1.35m. Rule 13 (Table 11.13)

: Add spacing for top and bottom vapor and liquid return

The tray spacing is assumed to be 0.55m. H = (tray spacing x Nactual ) + 3m H = (0.55m x 5.66) + 3m 𝐇 = 𝟔. 𝟏𝟏𝟑𝐦 Rule 14 (Table 11.13)

: Limit tower height to 53 m or L/D<30 𝐋 𝟔. 𝟏𝟏𝟑𝐦 = = 𝟒. 𝟓𝟐𝟖𝟏 < 𝟑𝟎 𝐃 𝟏. 𝟑𝟓𝐦

8.1.7 Separator 4 – Distillation Column Unit 2 Distillation column 2 is used for separating cumene. Thus, it is also known as Cumene column. Rule 2 (Table 11.13) : Relative volatility αavg =

P1sat P2sat


B T(˚C) + C

86

PLANT DESIGN FOR PRODUCTION OF PHENOL For Feed: Vapor pressure of Cumene at 100˚C, P1sat : log10 P1sat = 6.93666 −

1460.793 100 + 207.777

P1sat = 155.0205 mmHg P1sat = 0.20397 atm Vapor pressure of α– methylstyrene at 100˚C, P2sat : log10 P2sat = 7.0924 −

1582.7 100 + 206.01


P1sat 0.20397 atm = P2sat 0.10953 atm

𝛂𝐅𝐢𝐣 = 𝟏. 𝟖𝟔𝟐𝟐 For Distillate: Vapor pressure of Cumene at 165˚C, P1sat : log10 P1sat = 6.93666 −

1460.793 165 + 207.777


1582.7 165 + 206.01

P2sat = 670.6212mmHg P2sat = 0.88240 atm

87

PLANT DESIGN FOR PRODUCTION OF PHENOL Relative volatility for feed: αDij

P1sat 1.37141 atm = sat = 0.88240 atm P2 𝛂𝐃𝐢𝐣 = 𝟏. 𝟓𝟓𝟒𝟐

For Bottom: The relative volatility of the bottom part of the column is same as the distillate because its temperature condition is same which is 165˚C. Thus, 𝛂𝐁𝐢𝐣 = 𝛂𝐃𝐢𝐣 = 𝟏. 𝟓𝟓𝟒𝟐 The average volatility is: 3

αavg = √1.8622(1.5542)(1.5542) 𝛂𝐚𝐯𝐠 = 𝟏. 𝟔𝟓𝟎𝟕 Rule 7 (Table 11.13) : The minimum number of trays with Fenske-Underwood equation. For multi-component system using Fenske equation: ln [ Nmin = ln [ Nmin =


(0.7667/0.0026) ] (0.2029/0.0288) −1 ln 1.6507

𝐍𝐦𝐢𝐧 = 𝟔. 𝟒𝟓𝟎𝟔 Rule 6 (Table 11.13) : The optimum number of theoretical trays is near 2Nmin Nopt = 2Nmin = 12.9012 Rule 9 (Table 11.13) : A safety factor of 10% on number of trays Nactual = Nactual =

2Nmin (1 + 0.1) ε

12.9012(1 + 0.1) 0.85

𝐍𝐚𝐜𝐭𝐮𝐚𝐥 = 𝟏𝟔. 𝟔𝟗𝟓𝟕

88

PLANT DESIGN FOR PRODUCTION OF PHENOL Rule 4 (Table 11.14) : Tray efficiencies for distillation of light hydrocarbons and aq. Solution ε = 60 → 90% It is assumed that the tray efficiencies is 𝛆 = 𝟖𝟓%

Rule 8 (Table 11.13) : Minimum reflux at bubble point

R min

kmol kmol (72.4100 /18.0046 ) (F/D) hr hr = = (α − 1) (1.6507 − 1) 𝐑 𝐦𝐢𝐧 = 𝟔. 𝟏𝟖𝟎𝟕

Rule 5 (Table 11.13) : The optimum reflux in the range of 1.2-1.5R min R = (1.2 → 1.5)R min The optimum reflux is assumed to be 1.35𝐑 𝐦𝐢𝐧 R = (1.35)R min = 1.35(6.1807) 𝐑 = 𝟖. 𝟑𝟒𝟒𝟎

Rule 2 (Table 11.14) : Peak efficiency of vapor factor in range 1.2-1.5

m s

[kg/m3 ]0.5


𝐦 𝐬

[𝐤𝐠/𝐦𝟑 ]𝟎.𝟓

In order to find the density of the vapor in the distillate, equation below is used: ρn = ∑(xi ρi )n

89

PLANT DESIGN FOR PRODUCTION OF PHENOL For substance at distillate: Substances

Molar

Molecular

Mass

Mass

Density,

flowrate

weight

flowrate

fraction,

ρ (kg/m3)

(kmol/hr

(kg/kmol)

(kg/hr)

x

xρ

) Cumene

13.8046

120.19 1659.1749

Acetone

0.5460

58.08

α–

3.6540

118.18

0.7816

862.00

673.7392

31.7117

0.0149

784.00

11.6816

431.8297

0.2034

910.00

185.094

2122.7163

1.0000

methylstyrene Total

870.5148

ρv = 870.5148 kg/m3 u=

Fs 1.35 = 0.5 ρv 870.51480.5

𝐮 = 𝟎. 𝟎𝟒𝟓𝟖 𝐦/𝐬 The diameter, D of the tower is assumed to be 1.35m.

Rule 13 (Table 11.13)


The tray spacing is assumed to be 0.55m. H = (tray spacing x Nactual ) + 3m H = (0.55m x 16.6957) + 3m 𝐇 = 𝟏𝟐. 𝟏𝟖𝟐𝟔𝐦


: Limit tower height to 53 m or L/D<30 𝐋 𝟏𝟐. 𝟏𝟖𝟐𝟔𝐦 = = 𝟗. 𝟎𝟐𝟒𝟏 < 𝟑𝟎 𝐃 𝟏. 𝟑𝟓𝐦

90

PLANT DESIGN FOR PRODUCTION OF PHENOL 8.1.8 Separator 5 – Distillation Column Unit 3 Distillation column 4 is used for separating cumene. Thus, it is also known as Cumene column. Rule 2 (Table 11.13) : Relative volatility αavg =

P1sat P2sat


B T(˚C) + C

For Feed: Vapor pressure of cumene at 150˚C, P1sat : log10 P1sat = 6.93666 −

1460.793 150 + 207.777


1582.7 150 + 206.01


P1sat 0.94426 atm = P2sat 0.58334 atm

𝛂𝐅𝐢𝐣 = 𝟏. 𝟔𝟏𝟖𝟕 𝐚𝐭𝐦

91

PLANT DESIGN FOR PRODUCTION OF PHENOL For Distillate: Vapor pressure of cumene at 152˚C, P1sat : log10 P1sat = 6.93666 −

1460.793 152 + 207.777


1582.7 152 + 206.01

P2sat = 469.4332 mmHg P2sat = 0.61768 atm Relative volatility for distillate: αDij =

P1sat 0.98986 atm = P2sat 0.61768 atm

𝛂𝐃𝐢𝐣 = 𝟏. 𝟔𝟎𝟐𝟓 𝐚𝐭𝐦 For Bottom: The relative volatility of the bottom part of the column is same as the distillate because its temperature condition is same which is 152˚C. Thus, 𝛂𝐁𝐢𝐣 = 𝛂𝐃𝐢𝐣 = 𝟏. 𝟔𝟎𝟐𝟓 The average volatility is: 3

αavg = √1.6187(1.6025)(1.6025) 𝛂𝐚𝐯𝐠 = 𝟏. 𝟔𝟎𝟕𝟗 Rule 7 (Table 11.13) : The minimum number of trays with Fenske-Underwood equation. For multi-component system using Fenske equation: ln [ Nmin =


92


ln [ Nmin =

(0.7649/0.0451) ] (0.2045/0.9549 −1 ln 1.6079

𝐍𝐦𝐢𝐧 = 𝟖. 𝟐𝟎𝟓𝟒 𝐭𝐫𝐚𝐲 Rule 6 (Table 11.13) : The optimum number of theoretical trays is near 2Nmin Nopt = 2Nmin = 16.4108 tray Rule 9 (Table 11.13) : A safety factor of 10% on number of trays Nactual =

2Nmin (1 + 0.1) ε

Nactual =

16.4108(1 + 0.1) 0.85

𝐍𝐚𝐜𝐭𝐮𝐚𝐥 = 𝟐𝟏. 𝟐𝟑𝟕𝟔 𝐭𝐫𝐚𝐲 Rule 4 (Table 11.14) : Tray efficiencies for distillation of light hydrocarbons and aq. Solution ε = 60 → 90% It is assumed that the tray efficiencies is 𝛆 = 𝟖𝟓%


R min

kmol kmol (18.0046 /17.8666 ) (F/D) hr hr = = (α − 1) (1.6079 − 1)

𝐑 𝐦𝐢𝐧 = 𝟏. 𝟔𝟓𝟕𝟕

Rule 5 (Table 11.13) : The optimum reflux in the range of 1.2-1.5R min R = (1.2 → 1.5)R min The optimum reflux is assumed to be 1.35𝐑 𝐦𝐢𝐧 R = (1.35)R min = 1.35(1.6577) 𝐑 = 𝟐. 𝟐𝟑𝟕𝟗

93

PLANT DESIGN FOR PRODUCTION OF PHENOL Rule 2 (Table 11.14) : Peak efficiency of vapor factor in range 1.2-1.5

m s

94

[kg/m3 ]0.5


𝐦 𝐬

[𝐤𝐠/𝐦𝟑 ]𝟎.𝟓


Molar

Molecular

Mass

Mass

Density,

flowrate

weight

flowrate

fraction,

ρ (kg/m3)

(kmol/hr)

(kg/kmol)

(kg/hr)

x

Cumene

13.6666

α– methylstyrene

xρ

120.1900 1642.5887

0.7799

862.0000

672.2738

3.6540

118.1800

431.8297

0.2050

910.0000

186.5500

Acetone

0.5460

58.08

31.7112

0.0151

784.0000

11.8384

Total

5.3580

2106.1296

1.0000

ρv = 818.7084kg/m3 u=

Fs 1.35 = 0.5 ρv 870.66220.5

𝐮 = 𝟎. 𝟎𝟒𝟓𝟕𝟓 𝐦/𝐬 The diameter, D of the tower is assumed to be 1.35m.



The tray spacing is assumed to be 0.45m. H = (tray spacing x Nactual ) + 3m H = (0.55m x 21.2376) + 3m 𝐇 = 𝟏𝟒. 𝟔𝟖𝟎𝟕 𝐦

870.6622

PLANT DESIGN FOR PRODUCTION OF PHENOL Rule 14 (Table 11.13)

: Limit tower height to 53 m or L/D<30 𝐋 𝟏𝟒. 𝟔𝟖𝟎𝟕 𝐦 = = 𝟏𝟎. 𝟖𝟕𝟒𝟔 < 𝟑𝟎 𝐃 𝟏. 𝟑𝟓𝐦

8.1.9 Separator 6 – Distillation Column Unit 4 Distillation column 4 is used for separating cumene. Thus, it is also known as Cumene column. Rule 2 (Table 11.13) : Relative volatility αavg =

P1sat P2sat


B T(˚C) + C

For Feed: Vapor pressure of cumene at 152˚C, P1sat : log10 P1sat = 6.93666 −

1460.793 152 + 207.777


1582.7 152 + 206.01

P2sat = 469.433 mmHg P2sat = 0.61768 atm

95

PLANT DESIGN FOR PRODUCTION OF PHENOL Relative volatility for feed: P1sat 0.98995 atm = sat = 0.61768 atm P2

αFij

𝛂𝐅𝐢𝐣 = 𝟏. 𝟔𝟎𝟐𝟕 For Distillate: Vapor pressure of cumene at 165˚C, P1sat : log10 P1sat = 6.93666 −

1460.793 165 + 207.777


1582.7 165 + 206.01

P2sat = 670.06212 mmHg P2sat = 0.88166 atm Relative volatility for distillate: αDij

P1sat 1.37141 atm = sat = 0.88166 atm P2

𝛂𝐃𝐢𝐣 = 𝟏. 𝟓𝟓𝟓𝟔 For Bottom: The relative volatility of the bottom part of the column is same as the distillate because its temperature condition is same which is 165˚C. Thus, 𝛂𝐁𝐢𝐣 = 𝛂𝐃𝐢𝐣 = 𝟏. 𝟓𝟓𝟓𝟔 The average volatility is: 3

αavg = √1.6027(1.5556)(1.5556) 𝛂𝐚𝐯𝐠 = 𝟏. 𝟓𝟕𝟏𝟏

96

PLANT DESIGN FOR PRODUCTION OF PHENOL Rule 7 (Table 11.13) : The minimum number of trays with Fenske-Underwood equation. For multi-component system using Fenske equation: ln [ Nmin = ln [ Nmin =


(0.1901/0.7890) ] (0.0508/0.2110 −1 ln 1.3677

𝐍𝐦𝐢𝐧 = 𝟏. 𝟐𝟏𝟓𝟏 𝐭𝐫𝐚𝐲 Rule 6 (Table 11.13) : The optimum number of theoretical trays is near 2Nmin Nopt = 2Nmin = 2.4302 tray Rule 9 (Table 11.13) : A safety factor of 10% on number of trays Nactual =

2Nmin (1 + 0.1) ε

Nactual =

2.4302(1 + 0.1) 0.85

𝐍𝐚𝐜𝐭𝐮𝐚𝐥 = 𝟑. 𝟏𝟒𝟓𝟎 𝐭𝐫𝐚𝐲 Rule 4 (Table 11.14) : Tray efficiencies for distillation of light hydrocarbons and aq. Solution ε = 60 → 90% It is assumed that the tray efficiencies is 𝛆 = 𝟖𝟓%


R min


𝐑 𝐦𝐢𝐧 = 𝟓. 𝟖𝟑𝟖𝟖

97


98

Rule 5 (Table 11.13) : The optimum reflux in the range of 1.2-1.5R min R = (1.2 → 1.5)R min The optimum reflux is assumed to be 1.35𝐑 𝐦𝐢𝐧 R = (1.35)R min = 1.35(5.8388) 𝐑 = 𝟕. 𝟖𝟖𝟐𝟒 Rule 2 (Table 11.14) : Peak efficiency of vapor factor in range 1.2-1.5

m s

[kg/m3 ]0.5


𝐦 𝐬

[𝐤𝐠/𝐦𝟑 ]𝟎.𝟓

In order to find the density of the vapor in the distillate, equation below is used: ρn = ∑(xi ρi )n

For substance at distillate: Substances

xρ

Molar

Molecular

Mass

Mass

Density,

flowrate

weight

flowrate

fraction,

ρ (kg/m3)

(kmol/hr)

(kg/kmol)

(kg/hr)

x

Cumene

0.1367

120.1900

16.4300

0.3132

862.0000

269.9784

α– methylstyrene

0.0365

118.1800

4.3136

0.0822

910.0000

74.8020

Acetone

0.5460

58.08

31.7112

0.6045

784.0000

473.9280

Total

5.3580

52.4548

1.0000

ρv = 818.7084kg/m3 u=

Fs 1.35 = ρ0.5 818.70840.5 v

𝐮 = 𝟎. 𝟎𝟒𝟕𝟏𝟖 𝐦/𝐬 The diameter, D of the tower is assumed to be 1.35m.

818.7084

PLANT DESIGN FOR PRODUCTION OF PHENOL Rule 13 (Table 11.13)


The tray spacing is assumed to be 0.45m. H = (tray spacing x Nactual ) + 3m H = (0.55m x 3.1450) + 3m 𝐇 = 𝟒. 𝟕𝟑𝟎𝟎 𝐦 Rule 14 (Table 11.13)

: Limit tower height to 53 m or L/D<30 𝐋 𝟒. 𝟕𝟑𝟎𝟎 𝐦 = = 𝟑. 𝟓𝟎𝟑𝟕 < 𝟑𝟎 𝐃 𝟏. 𝟑𝟓𝐦

8.1.10 Separator 7 – Distillation Column Unit 5 Distillation column 5 is used for separating phenol. Thus, it is also known as Phenol column. Rule 2 (Table 11.13) : Relative volatility αavg =

P1sat P2sat


B T(˚C) + C

For Feed: Vapor pressure of α– methylstyrene at 175˚C, P1sat : log10 P1sat = 7.0924 −

1582.7 175 + 206.01

P1sat = 867.8422 mmHg P1sat = 1.1419 atm

99

PLANT DESIGN FOR PRODUCTION OF PHENOL Vapor pressure of Phenol at 175˚C, P2sat : log10 P2sat = 7.1345 −

1516.07 175 + 174.57


P1sat 1.1419 atm = P2sat 0.8255 atm

𝛂𝐅𝐢𝐣 = 𝟏. 𝟑𝟖𝟑𝟑 For Distillate: Vapor pressure of α– methylstyrene at 180˚C, P1sat : log10 P1sat = 7.0924 −

1582.7 180 + 206.01

P1sat = 982.3068 mmHg P1sat = 1.2925 atm Vapor pressure of Phenol at 180˚C, P2sat : log10 P2sat = 7.1345 −

1516.07 180 + 174.57

P2sat = 722.2726 mmHg P2sat = 0.9504 atm Relative volatility for feed: αDij

P1sat 1.2925 atm = sat = 0.9504 atm P2

𝛂𝐃𝐢𝐣 = 𝟏. 𝟑𝟔𝟎𝟎

100

PLANT DESIGN FOR PRODUCTION OF PHENOL For Bottom: The relative volatility of the bottom part of the column is same as the distillate because its temperature condition is same which is 180˚C. Thus, 𝛂𝐁𝐢𝐣 = 𝛂𝐃𝐢𝐣 = 𝟏. 𝟑𝟔𝟎𝟎 The average volatility is: 3

αavg = √1.3833(1.3600)(1.3600) 𝛂𝐚𝐯𝐠 = 𝟏. 𝟑𝟔𝟕𝟕 Rule 7 (Table 11.13) : The minimum number of trays with Fenske-Underwood equation. For multi-component system using Fenske equation: ln [ Nmin = ln [ Nmin =


(0.9147/0.0003) ] (0.0030/0.9618 −1 ln 1.3677

𝐍𝐦𝐢𝐧 = 𝟒𝟑. 𝟎𝟒𝟖𝟎 𝐭𝐫𝐚𝐲 Rule 6 (Table 11.13) : The optimum number of theoretical trays is near 2Nmin Nopt = 2Nmin = 86.0959 tray Rule 9 (Table 11.13) : A safety factor of 10% on number of trays Nactual =

2Nmin (1 + 0.1) ε

Nactual =

86.0959(1 + 0.1) 0.85

𝐍𝐚𝐜𝐭𝐮𝐚𝐥 = 𝟏𝟏𝟏. 𝟒𝟏𝟖𝟐 𝐭𝐫𝐚𝐲 Rule 4 (Table 11.14) : Tray efficiencies for distillation of light hydrocarbons and aq. Solution ε = 60 → 90% It is assumed that the tray efficiencies is 𝛆 = 𝟖𝟓%

101


102


R min


𝐑 𝐦𝐢𝐧 = 𝟖𝟕. 𝟑𝟎𝟑𝟐 Rule 5 (Table 11.13) : The optimum reflux in the range of 1.2-1.5R min R = (1.2 → 1.5)R min The optimum reflux is assumed to be 1.35𝐑 𝐦𝐢𝐧 R = (1.35)R min = 1.35(87.3032) 𝐑 = 𝟏𝟏𝟕. 𝟖𝟓𝟗𝟑 Rule 2 (Table 11.14) : Peak efficiency of vapor factor in range 1.2-1.5

m s

[kg/m3 ]0.5


𝐦 𝐬

[𝐤𝐠/𝐦𝟑 ]𝟎.𝟓


xρ

Molar

Molecular

Mass

Mass

Density,

flowrate

weight

flowrate

fraction,

ρ (kg/m3)

(kmol/hr)

(kg/kmol)

(kg/hr)

x

Cumene

0.1394

120.1900

16.7545

0.0836

862.0000

72.0390

α– methylstyrene

1.5503

118.1800

183.2145

0.9139

910.0000

831.6311

Phenol

0.0051

100.1500

0.5108

0.0025

1.070000

0.0027

Total

1.6948

200.4797

1.0000

903.6729

PLANT DESIGN FOR PRODUCTION OF PHENOL ρv = 903.6729 kg/m3 u=

Fs 1.35 = ρ0.5 903.67290.5 v

𝐮 = 𝟎. 𝟎𝟒𝟒𝟗 𝐦/𝐬 The diameter, D of the tower is assumed to be 1.8m.



The tray spacing is assumed to be 0.45m. H = (tray spacing x Nactual ) + 3m H = (0.45m x 111.4182) + 3m 𝐇 = 𝟓𝟑. 𝟏𝟑𝟖𝟐 𝐦


: Limit tower height to 53 m or L/D<30 𝐋 𝟓𝟑. 𝟏𝟑𝟖𝟐 𝐦 = = 𝟐𝟗. 𝟓𝟐𝟏𝟐 < 𝟑𝟎 𝐃 𝟏. 𝟖𝐦

103


9.0

COSTING OF EQUIPMENTS

9.1

Reactor 1 – Oxidation Reactor CSTR

From figure A.16, for V = 2.93 m3 Cpo = 7663 For FBM from table A.17, Reactor jacketed and agitated, FBM = 4 CBM = 7663 × 4 = 30 652 CBM,2017 = 30 652 ×

567.5 394

CBM = 𝟒𝟒 𝟏𝟒𝟗. 𝟕𝟕

9.2

Reactor 2 – Cleavage Unit CSTR

From figure A.16, for V = 29.07 m3 Cpo = 2714.86 For FBM from table A.17, Reactor jacketed and agitated, FBM = 4 CBM = 2714.86 × 4 = 10 859.43 CBM,2017 = 10 859.43 ×

567.5 394

CBM = 𝟏𝟓 𝟔𝟒𝟏. 𝟒𝟒

9.3

Reactor 3 – Pre-Purification Reactor CSTR

From figure A.16, for V = 0.1 m3 Cpo = 37 000 For FBM from table A.17, Reactor jacketed and agitated, FBM = 4 CBM = 37 000 × 4 = 148 000 CBM,2017 = 148 000 × CBM = 𝟐𝟏𝟑 𝟏𝟕𝟐. 𝟓𝟗

567.5 394

104


9.4

Separator 1 – Phase Separator (Gravity Settler)

Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5.

𝐕𝐨𝐥𝐮𝐦𝐞 =

πD2 L π(0.56 m)2 (1.4 m) = = 𝟎. 𝟑𝟒 𝐦𝟑 4 4

From Table A.1, Equipment data cost data for vertical process vessels is;

K1

3.4974

K2

0.4485

K3

0.1074

From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 3.4974 + 0.4485 log10 (0.34) + 0.1074 [log10 (0.34)]2 log10 C°p (2001) = 3.4974 − 0.2101 + 0.0236 C°p (2001) = 103.3109 = $ 2,045.97 The CEPCI for year 2017 is to be found as 567.5, so below is the calculation for 𝐂°𝐩 (2017): 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 2,045.97

567.5 = $ 𝟐, 𝟗𝟒𝟔. 𝟗𝟐 394

Table A.3 and Table A.4 shows,

B1

2.25

B2

1.82

FM

1.00

From Equation A.3,

𝐅𝐁𝐌 = 𝐁𝟏 + 𝐁𝟐 𝐅𝐩 𝐅𝐌

105

PLANT DESIGN FOR PRODUCTION OF PHENOL From Equation A.2,

𝐅𝐩,𝐯𝐞𝐬𝐬𝐞𝐥

Fp,vessel

(𝐏 + 𝟏)𝐃 + 𝟎. 𝟎𝟎𝟑𝟏𝟓 𝟐[𝟖𝟓𝟎 − 𝟎. 𝟔(𝐏 + 𝟏)] = 𝟎. 𝟎𝟎𝟔𝟑

(6.01325 + 1)(0.56) + 0.00315 2[850 − 0.6(6.01325 + 1)] = 0.0063

𝐅𝐩,𝐯𝐞𝐬𝐬𝐞𝐥 =

0.00232 = 0.3685 0.0063

Thus, 𝐅𝐁𝐌 = 2.25 + 1.82(0.3685)(1) = 2.9207 To obtain the Bare Module Cost for this phase separator, Equation 7.6 is used as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝐅𝐁𝐌 𝐂𝐁𝐌 = ($ 2,946.92)(2.9207) = $ 𝟖, 𝟔𝟎𝟕. 𝟎𝟕

106


9.5

Separator 2 – Scrubber

Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5.


πD2 L π(0.56 m)2 (1.4 m) = = 𝟎. 𝟑𝟒 𝐦𝟑 4 4

From Table A.1, Equipment data cost data for vertical process vessels is;

K1

3.4974

K2

0.4485

K3

0.1074

From Equation A.1,

log10Cop 2001  K1  K 2log10 A   K 3 log10 A 

2

log10C0p 2001  3.4974  0.4485log 10 0.34   0.1074log10 0.34 

2

Cop (2001)  10 3.3109 Cop (2001)  $2,045.97 The CEPCI for year 2017 is to be found as 567.5, so below is the calculation for 𝐂°𝐩 (2017):  567.5  Cop (2017)  $2,045.97    $2946.92  394 

Table A.3 and Table A.4 shows,

B1

2.25

B2

1.82

FM

1.00

From Equation A.3, FBM  B1  B2 Fp FM

107


(P  1)D  0.00315 2[850  0.6(P  1)] Fp,vessel  0.0063 (6.01325  1)(0.56)  0.00315 2[850  0.6(6.0135  1)] Fp,vessel  0.0063 0.00232 Fp,vessel   0.3685 0.0063 Thus,

FBM  2.25  1.82(0.368 5)(1)  2.9207

To obtain the Bare Module Cost for this phase separator, Equation 7.6 is used as following;

C BM  C op FBM C BM  ($2946.92) (2.9207) C BM  $8607.07

108


Separator 3 – Distillation Column Unit 1

9.6

For Tower: Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5. 𝐕𝐨𝐥𝐮𝐦𝐞 =

πD2 L π(1.35 m)2 (6.113 m) = = 𝟖. 𝟕𝟓𝟎𝟏 𝐦𝟑 4 4

From Table A.1, Equipment data cost data for towers tray and packed is;

K1

3.4974

K2

0.4485

K3

0.1074

From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 3.4974 + 0.4485 log10 (8.7501) + 0.1074 [log10 (8.7501)]2 log10 C°p (2001) = 3.4974 + 0.4225 + 0.0953 C°p (2001) = 104.0152 = $ 10,356.19 The CEPCI for year 2017 is to be found as 567.5, so below is the calculation for 𝐂°𝐩 (2017): 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 10,356.19

567.5 = $ 𝟏𝟒, 𝟗𝟏𝟔. 𝟓𝟗 394

Table A.3 and Table A.4 shows, For vertical carbon steel (CS) process vessels, the identification number is 18. B1

2.25

B2

1.82

FM

1.00

From Equation A.3,


109



Fp,vessel

(𝐏 + 𝟏)𝐃 + 𝟎. 𝟎𝟎𝟑𝟏𝟓 𝟐[𝟖𝟓𝟎 − 𝟎. 𝟔(𝐏 + 𝟏)] = 𝟎. 𝟎𝟎𝟔𝟑 (1.01325 + 1)(1.35) + 0.00315 2[850 − 0.6(1.01325 + 1)] = 0.0063


0.00475 = 0.7541 0.0063

Thus, 𝐅𝐁𝐌 = 2.25 + 1.82(0.7541)(1) = 3.6225 To obtain the Bare Module Cost for this phase separator, Equation 7.6 is used as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝐅𝐁𝐌 𝐂𝐁𝐌,𝐭𝐨𝐰𝐞𝐫 = ($ 14, 916.59)(3.6225) = $ 𝟓𝟒, 𝟎𝟑𝟓. 𝟑𝟓

For Tray: Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5. .


πD2 π(1.35 m)2 = = 𝟏. 𝟒𝟑𝟏𝟒 𝐦𝟐 4 4

From Table A.1, Equipment data cost data for sieve trays is;

K1

2.9949

K2

0.4465

K3

0.3961

110

PLANT DESIGN FOR PRODUCTION OF PHENOL From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 2.9949 + 0.4465 log10 (1.4314) + 0.3961 [log10 (1.4314)]2 log10 C°p (2001) = 2.9949 + 0.0695 + 0.0096 C°p (2001) = 103.074 = $ 1,185.77 The CEPCI for year 2017 is to be found as 567.5, so below is the calculation for 𝐂°𝐩 (2017): 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 1,185.77

567.5 = $ 𝟏, 𝟕𝟎𝟕. 𝟗𝟑 394

From Table A.5 shows, For sieve trays with N<20 due to N=5.66: 𝐥𝐨𝐠 𝟏𝟎 𝐅𝐪 = 𝟎. 𝟒𝟕𝟕𝟏 + 𝟎. 𝟎𝟖𝟓𝟏𝟔𝐥𝐨𝐠 𝟏𝟎 (𝐍) + 𝟎. 𝟑𝟒𝟕𝟑[𝐥𝐨𝐠 𝟏𝟎 (𝐍)]𝟐 log10 Fq = 0.4771 + 0.08516log10 (5.66) + 0.3473[log10 (5.66)]2 log10 Fq = 0.4771 + 0.0641 + 0.1968 𝐅𝐪 = 100.738 = 𝟓. 𝟒𝟕𝟎𝟐 From Table A.6, For stainless steel (SS) trays, the identification number is 61. 𝐅𝐁𝐌 = 𝟏. 𝟖

To obtain the Bare Module Cost for trays, Equation from Table A.5 is used for sieve trays as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝑵𝐅𝐁𝐌 𝑭𝒒 𝐂𝐁𝐌,𝐭𝐫𝐚𝐲 = ($ 1,707.93)(5.66)(1.8)(5.4702) = $ 𝟗𝟓, 𝟏𝟖𝟑. 𝟔𝟐

111


To obtain the Bare Module Cost for this distillation column, CBM = CBM,tower + CBM,tray CBM = $ 54,035.35 + $ 95,183.62 𝐂𝐁𝐌 = $ 𝟏𝟒𝟗, 𝟐𝟏𝟖. 𝟗𝟕


9.7

For Tower: Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5.


πD2 L π(1.35 m)2 (12.1826 m) = = 𝟏𝟕. 𝟒𝟑𝟖𝟎 𝐦𝟑 4 4


K1

3.4974

K2

0.4485

K3

0.1074

From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 3.4974 + 0.4485 log10 (17.4380) + 0.1074 [log10 (17.4380)]2 log10 C°p (2001) = 3.4974 + 0.5568 + 0.1655 C°p (2001) = 104.2197 = $ 16,584.41

112

PLANT DESIGN FOR PRODUCTION OF PHENOL The CEPCI for year 2017 is to be found as 567.5, so below is the calculation for 𝐂°𝐩 (2017): 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 16,584.41

567.5 = $ 𝟐𝟑, 𝟖𝟖𝟕. 𝟒𝟒 394


2.25

B2

1.82

FM

1.00

From Equation A.3,


From Equation A.2,


Fp,vessel

(𝐏 + 𝟏)𝐃 + 𝟎. 𝟎𝟎𝟑𝟏𝟓 𝟐[𝟖𝟓𝟎 − 𝟎. 𝟔(𝐏 + 𝟏)] = 𝟎. 𝟎𝟎𝟔𝟑 (1.01325 + 1)(1.35) + 0.00315 2[850 − 0.6(1.01325 + 1)] = 0.0063


0.00475 = 0.7541 0.0063

Thus, 𝐅𝐁𝐌 = 2.25 + 1.82(0.7541)(1) = 3.6225 To obtain the Bare Module Cost for this phase separator, Equation 7.6 is used as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝐅𝐁𝐌 𝐂𝐁𝐌,𝐭𝐨𝐰𝐞𝐫 = ($ 23,887.44)(3.6225) = $ 𝟖𝟔, 𝟓𝟑𝟐. 𝟐𝟓

113

PLANT DESIGN FOR PRODUCTION OF PHENOL For Tray: Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5. 𝐕𝐨𝐥𝐮𝐦𝐞 =

πD2 π(1.35 m)2 = = 𝟏. 𝟒𝟑𝟏𝟒 𝐦𝟐 4 4


K1

2.9949

K2

0.4465

K3

0.3961


567.5 = $ 𝟏, 𝟕𝟎𝟕. 𝟗𝟑 394

From Table A.5 shows, For sieve trays with N<20 due to N= 𝟏𝟔. 𝟔𝟗𝟓𝟕: 𝐥𝐨𝐠 𝟏𝟎 𝐅𝐪 = 𝟎. 𝟒𝟕𝟕𝟏 + 𝟎. 𝟎𝟖𝟓𝟏𝟔𝐥𝐨𝐠 𝟏𝟎 (𝐍) + 𝟎. 𝟑𝟒𝟕𝟑[𝐥𝐨𝐠 𝟏𝟎 (𝐍)]𝟐 log10 Fq = 0.4771 + 0.08516log10 (16.6957 ) + 0.3473[log10 (16.6957)]2 log10 Fq = 0.4771 + 0.1041 + 0.5191 𝐅𝐪 = 101.1003 = 𝟏𝟐. 𝟓𝟗𝟖𝟎

114

PLANT DESIGN FOR PRODUCTION OF PHENOL From Table A.6, For stainless steel (SS) trays, the identification number is 61. 𝐅𝐁𝐌 = 𝟏. 𝟖

To obtain the Bare Module Cost for trays, Equation from Table A.5 is used for sieve trays as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝑵𝐅𝐁𝐌 𝑭𝒒 𝐂𝐁𝐌,𝐭𝐫𝐚𝐲 = ($ 1,185.77)(16.6957 )(1.8)(12.5980) = $ 𝟒𝟒𝟖, 𝟗𝟑𝟎. 𝟓𝟗

To obtain the Bare Module Cost for this distillation column, CBM = CBM,tower + CBM,tray CBM = $ 86,532.25 + $ 448,930.59 𝐂𝐁𝐌 = $ 𝟓𝟑𝟓, 𝟒𝟔𝟐. 𝟖𝟒

9.8


For Tower: Based on Table 7.4, given CEPCI for year 2001 is 394 and 2017 is 567.5. πD2 L π(1.35 m)2 (14.6807 m) 𝐕𝐨𝐥𝐮𝐦𝐞 = = = 𝟐𝟏. 𝟎𝟏𝟑𝟖 𝐦𝟑 4 4 From Table A.1, Equipment data cost data for towers tray and packed is;

K1

3.4974

K2

0.4485

K3

0.1074

115

PLANT DESIGN FOR PRODUCTION OF PHENOL From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 3.4974 + 0.4485 log10 (21.0138 ) + 0.1074 [log10 (21.0138 )]2 log10 C°p (2001) = 3.4974 + 0.5931 + 0.1878 C°p (2001) = 104.2783 = $ 18,980.17 The CEPCI for year 2017 is to be found as 567.5, so below is the calculation for 𝐂°𝐩 (2017): 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 18,980.17

567.5 = $ 𝟐𝟕, 𝟑𝟑𝟖. 𝟏𝟖 394


2.25

B2

1.82

FM

1.00

From Equation A.3,


From Equation A.2,


(𝐏 + 𝟏)𝐃 + 𝟎. 𝟎𝟎𝟑𝟏𝟓 𝟐[𝟖𝟓𝟎 − 𝟎. 𝟔(𝐏 + 𝟏)] = 𝟎. 𝟎𝟎𝟔𝟑

Fp,vessel

(1.01325 + 1)(1.35) + 0.00315 2[850 − 0.6(1.01325 + 1)] = 0.0063


0.00475 = 0.7541 0.0063

116

PLANT DESIGN FOR PRODUCTION OF PHENOL Thus, 𝐅𝐁𝐌 = 2.25 + 1.82(0.7541)(1) = 3.6225 To obtain the Bare Module Cost for this phase separator, Equation 7.6 is used as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝐅𝐁𝐌 𝐂𝐁𝐌,𝐭𝐨𝐰𝐞𝐫 = ($ 27,338.18)(3.6225) = $ 𝟗𝟗, 𝟎𝟑𝟐𝟓. 𝟓𝟔

For Tray: Based on Table 7.4, given CEPCI for year 2001 is 394 and 2017 is 567.5.


πD2 π(1.35 m)2 = = 𝟏. 𝟒𝟑𝟏𝟒 𝐦𝟐 4 4


K1

2.9949

K2

0.4465

K3

0.3961


567.5 = $ 𝟏, 𝟕𝟎𝟕. 𝟗𝟑 394

117

PLANT DESIGN FOR PRODUCTION OF PHENOL From Table A.5 shows, For sieve trays with N>20 due to N= 𝟐𝟏. 𝟐𝟑𝟕𝟔 𝐅𝐪 = 𝟏 From Table A.6, For stainless steel (SS) trays, the identification number is 61. 𝐅𝐁𝐌 = 𝟏. 𝟖

To obtain the Bare Module Cost for trays, Equation from Table A.5 is used for sieve trays as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝑵𝐅𝐁𝐌 𝑭𝒒 𝐂𝐁𝐌,𝐭𝐫𝐚𝐲 = ($ 1,707.93)(21.2376 )(1.8)(1) = $ 𝟔𝟓, 𝟐𝟗𝟎. 𝟐𝟎

To obtain the Bare Module Cost for this distillation column, CBM = CBM,tower + CBM,tray CBM = $ 99,0325.56 + $ 65,290.20 𝐂𝐁𝐌 = $ 𝟏, 𝟎𝟓𝟓, 𝟔𝟏𝟓. 𝟕𝟔

118


9.9


For Tower: Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5. 𝐕𝐨𝐥𝐮𝐦𝐞 =

πD2 L π(1.35 m)2 (4.7300 m) = = 𝟔. 𝟕𝟕𝟎𝟓 𝐦𝟑 4 4


K1

3.4974

K2

0.4485

K3

0.1074


567.5 = $ 𝟏𝟐, 𝟔𝟔𝟏. 𝟎𝟓 394


2.25

B2

1.82

FM

1.00

From Equation A.3,


119



Fp,vessel

(𝐏 + 𝟏)𝐃 + 𝟎. 𝟎𝟎𝟑𝟏𝟓 𝟐[𝟖𝟓𝟎 − 𝟎. 𝟔(𝐏 + 𝟏)] = 𝟎. 𝟎𝟎𝟔𝟑

(1.01325 + 1)(1.35) + 0.00315 2[850 − 0.6(1.01325 + 1)] = 0.0063


0.00475 = 0.7541 0.0063

Thus, 𝐅𝐁𝐌 = 2.25 + 1.82(0.7541)(1) = 3.6225 To obtain the Bare Module Cost for this phase separator, Equation 7.6 is used as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝐅𝐁𝐌 𝐂𝐁𝐌,𝐭𝐨𝐰𝐞𝐫 = ($ 12,661.05)(3.6225) = $ 𝟒𝟓, 𝟖𝟔𝟑. 𝟔𝟓

For Tray: Based on Table 7.4, given CEPCI for year 2001 is 394 and CEPCI found for year 2017 is 567.5. πD2 π(1.35 m)2 𝐕𝐨𝐥𝐮𝐦𝐞 = = = 𝟏. 𝟒𝟑𝟏𝟒 𝐦𝟐 4 4 From Table A.1, Equipment data cost data for sieve trays is;

K1

2.9949

K2

0.4465

K3

0.3961

120

PLANT DESIGN FOR PRODUCTION OF PHENOL From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 2.9949 + 0.4465 log10 (1.4314) + 0.3961 [log10 (1.4314)]2 log10 C°p (2001) = 2.9949 + 0.0695 + 0.0096 C°p (2001) = 103.074 = $ 1,185.77 The CEPCI for year 2017 is to be found as 567.5, so below is the calculation for 𝐂°𝐩 (2017): 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 1,185.77

567.5 = $ 𝟏, 𝟕𝟎𝟕. 𝟗𝟑 394

From Table A.5 shows, For sieve trays with N<20 due to N=5.66: 𝐥𝐨𝐠 𝟏𝟎 𝐅𝐪 = 𝟎. 𝟒𝟕𝟕𝟏 + 𝟎. 𝟎𝟖𝟓𝟏𝟔𝐥𝐨𝐠 𝟏𝟎 (𝐍) + 𝟎. 𝟑𝟒𝟕𝟑[𝐥𝐨𝐠 𝟏𝟎 (𝐍)]𝟐 log10 Fq = 0.4771 + 0.08516log10 (3.1450 ) + 0.3473[log10 (3.1450)]2 log10 Fq = 0.4771 + 0.0424 + 0.0860 𝐅𝐪 = 100.6055 = 𝟒. 𝟎𝟑𝟏𝟖 From Table A.6, For stainless steel (SS) trays, the identification number is 61. 𝐅𝐁𝐌 = 𝟏. 𝟖

121

PLANT DESIGN FOR PRODUCTION OF PHENOL To obtain the Bare Module Cost for trays, Equation from Table A.5 is used for sieve trays as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝑵𝐅𝐁𝐌 𝑭𝒒 𝐂𝐁𝐌,𝐭𝐫𝐚𝐲 = ($ 1,707.93)(3.1450 )(1.8)(4.0318) = $ 𝟑𝟖, 𝟗𝟖𝟏. 𝟖𝟑

To obtain the Bare Module Cost for this distillation column, CBM = CBM,tower + CBM,tray CBM = $ 45,863.65 + $ 38,981.83 𝐂𝐁𝐌 = $ 𝟖𝟒, 𝟖𝟒𝟓. 𝟒𝟖

9.10 Separator 7 – Distillation Column Unit 5 For Tower: Based on Table 7.4, given CEPCI for year 2001 is 394 and 2017 is 567.5.

πD2 L π(1.8 m)2 (53.1382 m) 𝐕𝐨𝐥𝐮𝐦𝐞 = = = 𝟏𝟑𝟓. 𝟐𝟐𝟎𝟐 𝐦𝟑 4 4 From Table A.1, Equipment data cost data for towers tray and packed is;

K1

3.4974

K2

0.4485

K3

0.1074

122

PLANT DESIGN FOR PRODUCTION OF PHENOL From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 3.4974 + 0.4485 log10 (135.2202) + 0.1074 [log10 (135.2202)]2 log10 C°p (2001) = 3.4974 + 0.9558 + 0.4877 C°p (2001) = 104.9409 = $ 87 277.04 From Table 7.4, the latest year for Equipment Cost Index is year 2017. Hence, the purchased cost of vertical process vessels for year 2017 is shown below; 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 87 277.04

567.5 = $ 𝟏𝟐𝟓 𝟕𝟎𝟗. 𝟗𝟓 394


2.25

B2

1.82

FM

1.00

From Equation A.3,


From Equation A.2,


Fp,vessel

(𝐏 + 𝟏)𝐃 + 𝟎. 𝟎𝟎𝟑𝟏𝟓 𝟐[𝟖𝟓𝟎 − 𝟎. 𝟔(𝐏 + 𝟏)] = 𝟎. 𝟎𝟎𝟔𝟑 (1.01325 + 1)(1.8) + 0.00315 2[850 − 0.6(1.01325 + 1)] = 0.0063


0.00528 = 0.8381 0.0063

123

PLANT DESIGN FOR PRODUCTION OF PHENOL Thus, 𝐅𝐁𝐌 = 2.25 + 1.82(0.8381)(1) = 3.7753

To obtain the Bare Module Cost for this phase separator, Equation 7.6 is used as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝐅𝐁𝐌 𝐂𝐁𝐌,𝐭𝐨𝐰𝐞𝐫 = ($ 125 709.95)(3.7753) = $ 𝟒𝟕𝟒 𝟓𝟗𝟐. 𝟕𝟕

For Tray: Based on Table 7.4, given CEPCI for year 2001 is 394 and 2017 is 567.5.


πD2 π(1.8 m)2 = = 𝟐. 𝟓𝟒𝟒𝟕 𝐦𝟐 4 4


K1

2.9949

K2

0.4465

K3

0.3961

From Equation A.1, 𝐥𝐨𝐠 𝟏𝟎 𝐂°𝐩 (𝟐𝟎𝟎𝟏) = 𝐊 𝟏 + 𝐊 𝟐 𝐥𝐨𝐠 𝟏𝟎 (𝐀) + 𝐊 𝟑 [𝐥𝐨𝐠 𝟏𝟎 (𝐀)]𝟐 log10 C°p (2001) = 2.9949 + 0.4465 log10 (2.5447) + 0.3961 [log10 (2.5447)]2 log10 C°p (2001) = 2.9949 + 0.1811 + 0.0652 C°p (2001) = 103.2412 = $ 1742.61 From Table 7.4, the latest year for Equipment Cost Index is year 2017. Hence, the purchased cost of vertical process vessels for year 2017 is shown below; 𝐂°𝐩 (𝟐𝟎𝟏𝟏) = $ 1,742.61

567.5 = $ 𝟐 𝟓𝟎𝟗. 𝟗𝟖 394

124

PLANT DESIGN FOR PRODUCTION OF PHENOL From Table A.5 shows, For sieve trays with N>20 due to N=111.4182 trays: 𝐅𝐪 = 𝟏. 𝟎 From Table A.6, For stainless steel (SS) trays, the identification number is 61. 𝐅𝐁𝐌 = 𝟏. 𝟖 To obtain the Bare Module Cost for trays, Equation from Table A.5 is used for sieve trays as following; 𝐂𝐁𝐌 = 𝐂°𝐩 𝐍𝐅𝐁𝐌 𝐅𝐪 𝐂𝐁𝐌,𝐭𝐫𝐚𝐲 = ($ 2509.98)(111.4182)(1.8)(1.0) = $ 𝟓𝟎𝟑 𝟑𝟖𝟑. 𝟒𝟐 To obtain the Bare Module Cost for this distillation column, CBM = CBM,tower + CBM,tray CBM = $ 474 592.77 + $ 503 383.42 𝐂𝐁𝐌 = $ 𝟗𝟕𝟕 𝟗𝟕𝟔. 𝟏𝟗

125


9.11 Total Equipment Cost Equipments

No

Price ($)

1

Oxidation Reactor

44,149.77

2

Cleavage Unit Reactor

15,641.44

3

Pre-Purification Reactor

4

Phase Separator

8,607.07

5

Scrubber

8,607.07

6

Distillation Column 1

149,218.97

7


535,462.84

8


1,055,615.76

9


84,845.48

10


977,976.19

TOTAL

213,172.59

3,093,297.18

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10.0 OPERATING LABOUR COST A year consists of 52 weeks. A single operator will work on average of 49 weeks per year with 3 weeks off on 8 hours per shift and 5 shift per week. Usually a chemical plant operates for 24 hours so it requires 3 shift per day for 365 days. Thus, (49 weeks/year) x (5 shifts/week)

= 245 shifts/year

(365 days/year) x (3 shifts/day)

= 1095 shifts/year

(1095 shifts/year) x (Operator.year/245 shifts)

= 4.5 operators

Four and one-half operators are hired for each operator needed in the plant at any time. In order to find the value of number of operating labor required to run the process unit per shift, equation as stated below is used: 𝐍𝐎𝐋 = (𝟔. 𝟐𝟗 + 𝟑𝟏. 𝟕𝐏 𝟐 + 𝟎. 𝟐𝟑𝐍𝐧𝐩 )𝟎.𝟓 NOL

= Number of operating labor operating per shift

P2

= Particulate processing steps

Nnp

= Non-particulate processing steps

Table 20 Shows The Result For The Estimation Of Operating Labor Requirements For The Process Using The Equipment Module Approach Equipment Type

Number of Equipment

Nnp

Compressors

2

2

Heat Exchangers

12

12

Heaters/Furnaces

0

0

Pumps*

9

-

Reactors

3

3

Towers

6

6

Vessels*

5

-

Total

23

*Pumps and vessels are not counted in evaluating Nnp. P2 = 0 because there is no particulate processing steps in this process.

127

PLANT DESIGN FOR PRODUCTION OF PHENOL NOL = (6.29 + 31.7(0)2 + 0.23Nnp )0.5 NOL = (6.29 + 31.7(0)2 + 0.23(23))0.5 𝐍𝐎𝐋 = 𝟑. 𝟒𝟎 Number of operators required per shift is 3.40.

Number of operator needed as operating labor: 3.40 x 4.5 operators = 15.3 (rounding up yields 16 operators)

For all equipment, cost of operating labor per year: (RM 6.25/hour.operator) x (8 hours/day) x (365 days/year) x (16 Operator) = RM 292,000/year

128


12.0 CONCLUSION Phenol is the precursor to many materials and a very useful compounds. The global phenol market predicts to witness a robust CAGR of 6.8% from 2017 to 2022. The cumene process segment should be worth almost US$12.3 billion in 2017. The plant is to be built in Malaysia and three chosen state which is Kerteh (Terengganu), Gebeng (Pahang) and Pasir Gudang (Johor) is scored using Scoring Method to choose for the plant site. Pasir Gudang Industrial Estate, Johor is chosen for the plant site. The Block Flow Diagram (BFD) and Process Flow Diagram (PFD) is made based on the selected process of production. The process chosen for the plant operation is production of phenol from cumene peroxide using cumene as the main raw material. The equipment cost is predicted from the sizing which revealed the bare module cost of the equipment yielding to approximately $3,093,297.18. The operating labour cost approximately RM 292,000/year is estimated using the Equipment Module Approach. The plant is estimated to produced 51 kmol/h of phenol for every 100 knol/h of cumene over the plant start-up.

129


BIBLIOGRAPHY Brydson, J. (1999). Phenolic Resins. Polymer Matrtix Composites and Technology, 1. Chaplin, M. (6 August, 2014). Enzyme Technology. Retrieved from http://www1.lsbu.ac.uk/water/enztech/cstr.html Commodity, S. (10 October, 2018). Industrial Sector. Retrieved from Sunsirs Commodity Data Group: http://www.sunsirs.com/uk/sectors-14.html Elvers, B. (2011). Ullmann’s Encyclopedia of Industrial Chemistry. Wiley- VCH Verlag GmbH & Co. Elyers, B. (2011). Ullmann's Encyclopedia of Industrial Chemistry . Wiley-VCH Verlag GmbH & Co. Estate, G. I. (n.d.). Facilities and Infrastructure. Retrieved from Invest In Pahang: http://www.investinpahang.gov.my/index.php?rp=gebeng_facilities_infra.pdf Guide, E. (2007). Types of Phenol Manufacturing Process. Retrieved from Engineers Guide: http://enggyd.blogspot.com/2010/07/phenol-manufacturing-process.html Industrial Sector. (10 October, 2018). Retrieved from Sunsirs Commodity Data Group: http://www.sunsirs.com/uk/sectors-14.html Kantarci, N., Borak, F., & Ulgen, K. (2005). Review: bubble column reactors. Process Biochemistry , 2263-2283. Kennepohl, D. (2017). Phenol And Their Uses. Chemistry and Libretexts. Kirk-Othmer. (n.d.). Encyclopedia of Chemical Technology Fifth Edition. KualitiAlam. (n.d.). Retrieved from Kualiti Alam: www.kualitialam.com L. Ross C. Barclay, M. B. (2003). Chemistry of Phenol. England: John Wiley & Sons Ltd. Megson, N. J. (1958). Phenolic Resin Chemistry . London: Butterworths Scientific Publications. Mengmeng, S., & Ekaterina, S. (2017). Design of Bubble Column Reactor.

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PLANT DESIGN FOR PRODUCTION OF PHENOL MIDA. (2017). Malaysia Costs Of Doing Business. Malaysia: The Malaysian Investment Development Authority (MIDA). Payscale. (Dec, 2018). Average Salary in Johor Bahru, Malaysia. Retrieved from Payscale: https://www.payscale.com/research/MY/Location+JohorBahru/Hourly_Rate#by_Skill Sarina, S., Huaiyong, Z., Zhanfeng, Z., & Guoran, L. (2012). Driving selective aerobic oxidation of alkyl aromatics by sunlight on alcohol grafted metal hydroxides. Chemical Science, 2138-2146. Sayyar, M. H. (2008). Direct Oxidation of Benzene to Phenol . Schmidt, R. (2005). Industrial catalytic processes-phenol production. ScienceLab. (n.d.). Material Safety Data Sheet Listing. Retrieved from Science Lab: http://www.sciencelab.com/msdsList.php Solomon, M. G. (1995). Aust. J. Chem, 323. Tobiason, F. L. (1981). Phenolic Resins—Chemistry and Applications. Weyerhaeuser Science Symposium, 2, 201–212. Ullmann. (2005). Ullmann's Encyclopedia of Industrial Chemistry. In M. W.-B. Manfred Weber, Phenol. Germany: Wiley Publisher.

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CPE604 - Mini Project Plant Design (Production Of Phenol) (2018)

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