Vcm Production 100tons/day

Chapter1 Introduction ______________________________________________________________________________________

Vinyl chloride is an organic chloride with the formula H2C=CHCl that is also called vinyl chloride monomer (VCM) or chloroethene. This colorless compound is an important industrial chemical chiefly used to produce the polymer polyviny chloride (PVC).About 13 billion kilograms are produced annually. VCM is among the top twenty largest petrochemicals (petroleum-derived chemicals) in world production. The United States currently remains the largest VCM manufacturing region because of its lowproduction-cost position in chlorine and ethylene raw materials. China is also a large manufacturer and one of the largest consumers of VCM. Vinyl chloride is a gas with a sweet odor. Vinyl chloride that is released by industries or formed by the breakdown of other chlorinated chemicals can enter the air and drinking water supplies. Vinyl chloride is a common contaminant found near landfills. In the past VCM has been used as a refrigerant.

Vinyl Chloride Monomer (VCM) is the key material from which PVC is made. VCM is a gas with a molecular weight of 62.5 and boiling point of -13.9°C, and hence has a high vapour pressure at ambient temperature. It is therefore manufactured under strict quality and safety control. There are two ways to manufacture VCM from ethylene (obtained from thermal cracking); the direct chlorination method and oxychlorination method (Figure 1.1). VC is transported as a compressed liquid. As it does not tend to polymerize easily, liquid VCM(in the absence of oxygen and water) can be stored and transported without polymerization inhibitors (Bönnighausen, 1986).

Figure 1.1. Preparation of Vinyl Chloride Monomer by oxychlorination and cracking method ______________________________________________________________________________________ 1

Chapter1 Introduction ______________________________________________________________________________________

1.1 PROPERTIES Vinyl chloride is an organic colourless compound. The physical and chemical properties of Vinyl chloride are presented in Table 1.1. Table 1.1 Physical properties

Chemical formula

C2H3Cl

Molar mass

62.50 g·mol−1

Appearance

Colorless gas

Odor

Pleasant

Density

0.969 g/ml

Melting point

−153.8 °C (−244.8 °F; 119.3 K)

Boiling point

−13.4 °C (7.9 °F; 259.8 K)

Solubility in water

2.7 g/L (0.0432 mol/L)

Vapor pressure

2580 mm. of mercury 20 °C (68 °F)

Magnetic susceptibility (χ)

-35.9·10−6 cm3/mol

Specific heat capacity (C)

0.8592 J/K/g(gas) 0.9504 J/K/g (solid)

Std enthalpy (ΔHo298)

−94.12 kJ/mol (solid)

Critical Temperature

156.6˚C

Critical Pressure

5.6 MPa

Critical Volume

169 cm3/mol

Latent Heat of Fusion

75.9 J/g

Latent Heat of Vaporisation

330 J/g

Standard Gibbs energy of Formation

51.5 kJ/ mol

Flash Point

-77.75˚C

1.1.1 Flammability VCM is extremely flammable. At concentrations of about 3.6 percent VCM in air, VCM can be an explosion hazard. Direct contact with open flames or a high energy heat source will result in combustion and corrosive, noxious gases. If combustion occurs, extinguish fires using dry chemical, foam, or carbon dioxide. Water may be ineffective, but should be used to keep fire-exposed containers cool. 1.1.2 Reactivity VCM will polymerize if exposed to air, elevated temperatures or other activating substances. Inhibitors are often added to VCM to prevent polymerization during storage. VCM can be stored in vessels made of common materials of construction. VCM is stable ______________________________________________________________________________________ 2

Chapter1 Introduction ______________________________________________________________________________________

with common metals other than aluminum and aluminum alloys and copper and copper alloys (including brass). When moisture is present, VCM can corrode iron and steel. Avoid VCM contact with moisture, pure oxygen, strong alkalis, alkali metals, open flames and welding arcs, and other high temperature sources, which induce thermal decomposition to irritating and corrosive hydrochloric acid. 1.1.3 Solubility Water at 25 °C

---- 2,763 mg/L EPA 1,100 mg/L

Organic solvent(s) ---- Soluble in hydrocarbons, oil and most common organic solvents

1.2 APPLICATIONS Vinyl is versatile: it can be as rigid as industrial pipes, as pliable as plastic wrap, and as thin and flexible as wallcovering. It can also be completely clear or matched to any color desired. 1.2.1. Building and Construction About three-quarters of all vinyl produced goes into long-lasting building and construction applications. Life-cycle studies show PVC/vinyl is effective in protecting the environment, in terms of low greenhouse gas emissions and conservation of resources and energy. Because it is strong and resistant to moisture and abrasion, vinyl is ideal for cladding, windows, roofing, fencing, decking, wallcoverings, and flooring. Vinyl does not corrode like some building materials, does not require frequent painting and can be cleaned with mild cleaning products. 1.2.2 Siding and Windows Vinyl is used to produce siding and window frames that are extremely durable, affordable, and help to conserve energy during heating and cooling homes. In fact, vinyl windows have three times the heat insulation of aluminum windows. 1.2.3 Wiring and Cables Vinyl is able to withstand tough conditions behind building walls – such as exposure to changing temperatures and dampness – for the life of the building. As a result, it is one of the most prevalent and trusted materials used in electrical wiring and cables. 1.2.4 Water Pipes PVC helps conserve energy and water by creating virtually leak-free pipes that are not prone to corrosion and resist environmental stress. PVC breakage rates are as low as one

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Chapter1 Introduction ______________________________________________________________________________________

percent of the breakage rates of cast metal systems. The lack of build-up in PVC piping improves functionality and increases energy efficiency. 1.2.5 Packaging Because it is durable, dependable and light weight, flexible PVC helps packaging do its job to maintain the integrity of the products inside, including medicines. Clear vinyl is used in tamper-resistant over-the-counter medications and shrink wrap for consumer products. Rigid vinyl film is used in blister and clamshell packaging to protect medicines, personal care products and other household goods. 1.2.6 Healthcare Vinyl plays a critical safety role in dispensing life-saving medicine through IV bags and medical tubing. The advent of the PVC blood-collection bag was a significant breakthrough because blood bags are flexible and unbreakable, enhancing the development of ambulatory medicine and serving as the foundation for modern blood banks. 1.2.7 Household Products PVC’s affordability, durability and water resistance make it ideal for rain coats, boots and shower curtains.

1.3 MARKET SURVEY Vinyl chloride monomer (VCM) is one of the world’s most important and largest commodity chemicals. VCM is used primarily for the production of polyvinyl chloride (PVC) homopolymer and copolymer resins. PVC has the advantage of being utilized in conversion and fabrication processes with great flexibility, such that end products cover a wide range, including pipe and fittings, profiles and tubes, siding, wire and cable, windows, doors, floorings, film and sheet, and bottles. While the most important ultimate end-use markets are commercial, residential, and nonresidential construction, a wide variety of PVC converted products are also utilized in agricultural, electrical (wire and cable), and health care markets. The vinyls chain, comprising ethylene dichloride (EDC), vinyl chloride monomer (VCM), and polyvinyl chloride (PVC), is a key component of the global petrochemical and thermoplastics sectors. The vinyls industry—and VCM, as part of the vinyls chain—has a history of change; manufacturers have exited and/or consolidated, and new firms have been created over the decades. Manufacturing technology has been improved from the standpoint of safety, the environment, quality, and scale of production. ______________________________________________________________________________________ 4

Chapter1 Introduction ______________________________________________________________________________________

World consumption of vinyl chloride monomer is shown in Figure 1.2

Figure 1.2 World consumption of VCM

1.4 CONSUMPTION & DEMAND In 2016, Northeast Asia was the largest consumer of VCM, accounting for over half of the world’s VCM demand. China is the largest overall player in the VCM market, with 43% of the total global capacity and about 38% of total global production 2016. China also leads the world in VCM consumption, accounting for 40% of global demand in 2016; demand will continue to grow by 3.7% annually through the forecast period to 2021. The second-largest consumer was North America, representing 19% of global demand, with the United States being the main driver in the region. The United States is the second-largest overall player worldwide, and maintains a low production cost position in chlorine and ethylene raw materials. The movement toward lower natural gas and feedstock costs for the vinyls chain in the United States and Canada, via shale gas, is solidifying the North American position as one of the world’s lowest-cost VCM producers. In 2016, the total volume of VCM exported globally represented 7% of world production. The largest exporters are North America and Northeast Asia, which together account for 84% of the world’s VCM exports. Over the next five-year period, VCM trade is expected to decline by about 0.6% per year as more vinyls producers become integrated. VCM trade is challenging because of logistics, as VCM is a gas that needs to be shipped and stored liquefied under pressure, requiring expensive LPG ships for its transportation.

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Chapter1 Introduction ______________________________________________________________________________________

Over 99% of total global VCM consumption in 2016 was used for PVC production; VCM, therefore, follows the PVC market trend very closely. Demand for VCM is expected to grow at an average annual rate of about 3% during 2016–21. During the past few years, the global VCM market had been growing at nearly 4%/year. However, growth rates in the developed world had been slipping as these markets matured while stronger demand growth has been seen in Asia, in particular China and India.

1.5 LEADING PRODUCERS The first large scale production units for this route were constructed by Dow Chemical Co., Monsanto Chemical Co. and the Shell Oil Co. The complete changeover to the exclusive use of ethylene as a feedstock became possible when the large-scale oxychlorination of ethylene to 1,2-dichloroethane had been proven to be technically feasible by Dow Chemical in the period 1955 – 1958. Since then, most plants now are using integrated, balanced DC– EDC – Oxy – EDC –VCM processes and more than 90% of the vinyl chloride presently produced in the Western World is derived from ethylene. RIL, Finolex, Chemplast, DCW, Shriram and Vivanta are the major producers of PVC resin and related products in India.

1.6. OBJECTIVE AND SCOPE The project aims at designing a process for production of vinyl chloride from ethylene and chlorine. The objectives of the project are: 1. To design a setup for manufacturing Vinyl Chloride at 1000 tons/day. 2. To carry out the literature survey on the manufacture of Vinyl Chloride. 3. To select a suitable process to manufacture Vinyl Chloride 4. To carry out material and energy balance for the process. 5. To design the equipment involved in the process. 6. To evaluate the economic feasibility of the process. 7. To select the plant location and layout.

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Chapter2

Literature survey

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The literature survey of a compound consists of the different manufacturing processes. The specifications and advantages of the selected process are listed in the following chapter. Vinyl chloride was first produced using the process of dehydrating ethylene dichloride (EDC) with alcoholic caustic potash. However, the first effective industrial process was based on the hydro chlorination of acetylene. Until the late 1940s, this process was used almost exclusively. However, as ethylene became more plentiful in the early 50’s, commercial processes were developed to produce vinyl chloride from chlorine and ethylene via EDC, namely, the balanced ethylene route. All current production plants for vinyl chloride depend on the use of a C2 hydrocarbon feedstocks, specifically, acetylene, ethylene, or ethane. Commercial operations using these compounds are confined to gas-phase processes. 2.1 Different Manufacturing Process 2.1.1 Vinyl Chloride from Ethane Many attempts have been made to develop a process that will use ethane to directly produce vinyl chloride. This is due to relative inexpensiveness of ethane. The major problem associated with the use of ethane is its molecular symmetry. In particular, the addition of chlorine to ethane gives rise to a wide product spectrum. 2.1.2 Vinyl Chloride from Acetylene The process that produces vinyl chloride from acetylene employs the use of a catalyst. Most of the time the catalyst used is mercuric chloride deposited on active carbon. In this process the feed gases are purified, dried, and mixed at the entrance to the tubular fixed bed reactors, which are packed with mercuric chloride on active carbon pellets as catalysts. 2.1.3 Vinyl Chloride from Ethylene Ethylene can be converted to vinyl chloride in a single stage, i.e., without isolating the intermediate ethylene dichloride by either chlorination or oxychlorination routes, as is the case with the balanced ethylene route. Direct chlorination routes require a high temperature and a large excess of ethylene to minimize soot formation ______________________________________________________________________________________ 7

Chapter2

Literature survey

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2.2 BALANCED PROCESS OVERVIEW Direct chlorination : CH2CH2 + Cl2 → ClCH2CH2Cl

Eq(1)

Oxy chlorination : CH2CH2 + 2 HCl + ½ O2 → ClCH2CH2Cl + H2O

Eq(2)

EDC pyrolysis

: 2ClCH2CH2Cl → 2 CH2CHCl + 2 HCl

Eq(3)

Overall reaction

: 2CH2CH2 + Cl2 + ½ O2 → 2 CH2CHCl + H2O

Eq(4)

2.3 PROCESS SELECTION The five main processes used in the production of vinyl chloride monomer (VCM) are: (1) direct chlorination of ethylene to form EDC, (2) oxychlorination of ethylene to form EDC from recycled HCl and oxygen, (3) purification of EDC, (4) thermal cracking of EDC to form VCM and HCl, and (5) the purification of VCM. These process is shown in the Figure.2.1.

Figure. 2.1 Process

flow sheet for the selected process

Ethylene and Cl2 combine in a homogeneous catalytic reaction to form EDC. Normally, the reaction rate is controlled by mass transfer, with absorption of ethylene as the limiting factor. Due to high selectivity, ferric chloride is the common catalyst of choice for chlorination of ethylene. The catalytic reaction utilizes an electrophilic addition ______________________________________________________________________________________ 8

Chapter2

Literature survey

______________________________________________________________________________________

mechanism. The catalyst polarizes chlorine (Eq.5) and then the polarized chlorine molecule acts as an electrophilic reagent to add Cl to the double bond of ethylene (Eq. 6). FeCl3 + Cl2 ↔ FeCl4-Cl+

Eq (5)

FeCl4-Cl+ + CH2CH2 → FeCl3 + ClCH2CH2Cl

Eq(6)

The EDC product from then reactor is cooled to room temperature before sending it to the storage vessel and then cracked in pyrolysis chamber to form VCM. The unconverted EDC and formed VCM is sent to a quencher where the recycled EDC is added to stabilize the mixture before sent to distillation columns for further purification. Two distillation columns are used to separate vinyl chloride from EDC, HCl and remaining by-products. The first column, HCl column, distills the hydrogen chloride mixture to a pure overhead product. This HCl is recycled to the oxychlorination reactor. The HCl column operates at a top tray pressure of 135 psig with a column pressure drop of 10 psig. The bottoms product of the HCl column is fed to the second column, the VCM column. The VCM column operates at a top tray pressure of 65 psig with a column pressure drop of 10 psig. A VCM product of 99.9 wt% is produced as the overhead product of the VCM column. The bottoms of the VCM column are recycled to the lights column for re-purification.

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Chapter2

Literature survey

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Chapter 3 Thermodynamic feasibility ____________________________________________________________________________________

The feasibility of a process is calculated by determining the thermodynamic parameters. Gibbs free energy is a thermodynamic property that predicts whether a process will occur spontaneously at constant temperature and pressure. Gibbs free energy G is defined as G=H-TS where H, T and S are enthalpy, temperature and entropy. Change in Gibbs free energy G correspond to changes in free energy for processes at constant temperature and pressure. The change in Gibbs free energy change is the maximum non expansion work obtainable under these conditions. ΔG is negative for spontaneous processes, positive for non-spontaneous process and zero for processes at equilibrium. Thermodynamic data for all the components are given in Table 3.1. The reactions are: C2H4 + Cl2 C2H4Cl

50ᵒC, 2 atm 500ᵒC, 4atm

C2H4Cl2

( Direct Chlorination)

C2H3Cl + HCl

(Pyrolysis Furnace)

Table 3.1 Thermodynamic properties of the reactants and products

Component

Boiling Point (⁰C)

Molecular Weight (g/mol)

Enthalpy H (kJ/kmol) (50ᵒC ,2atm)

Enthalpy H (kJ/kmol) (500ᵒC ,4atm)

Gibbs Free Energy ΔG⁰ (kJ/kmol)

Ethylene

-103.9

28

53607.47

-

7505.35

Chlorine

-34.6

71

0

-

0

Ethylene dichloride Hydrogen Chloride

83.47

99

-127814.75

-79220.547

-8256.55

-85

36.5

-

-78378.64

-95300

Vinyl Chloride

-12

62.5

-

64473.75

41950

3.1. HEAT OF REACTION The amount of heat generated or consumed in a chemical reaction depends on the operating conditions. Heat of reaction can be calculated from ΔHReaction = ∑ ΔHProducts - ∑ΔHReactants 3.1.1. Direct Chlorination Reaction : CH2CH2 + Cl2 → ClCH2CH2Cl ΔHReaction1 = (-127814.75 ) – (53607.47) = -181422.22 kJ /kmol (Exothermic Reaction) ______________________________________________________________________________________ 11

Chapter 3 Thermodynamic feasibility ____________________________________________________________________________________

3.1.2. Pyrolysis Reaction: 2ClCH2CH2Cl → 2CH2CHCl + 2HCl ΔHReaction2 = 64473.75-(-78378.64 – 79220.547) = 65315.657 kJ /kmol (Endothermic Reaction)

3.2. FEASIBILITY Gibbs Free Energy predicts the feasibility and equilibrium condition at constant temperature and pressure. ΔGReaction = ∑ ΔGProducts - ∑ΔGReactants 3.2.1. Direct Chlorination ΔG⁰1

= -8256.55 - (7505.35) = -15465.82 kJ /kmol

K1

= 𝑒𝑥𝑝

ln

=

K2

= 1.7765

∆𝐺

= −𝑅𝑇 ln(𝐾 ) = −8.314 ∗ 323 ∗ ln(1.7765) = -1543.164 kJ /kmol (As ∆𝐺 is negative, reaction is feasible )

∆ °

∆

.

= 𝑒𝑥𝑝(−

∗

) = = 514.0539

.

=

−

.

−

.

3.2.2 Pyrolysis ∆𝐺

= (41950 -95300) - (-8256.55) = - 45439 kJ /kmol

K

=

𝑒𝑥𝑝(−

.

∗

)

= 𝐾 ln 𝐾 K2 ∆𝐺

92.26E+06 −65315.657 1 1 = ∗ − 8.314 298 773 = 1E+15 = −𝑅𝑇 ln(𝐾 ) = −8.314 ∗ 773 ∗ ln(1𝐸 + 15) = - 221971.11 kJ /kmol (As ∆𝐺 is negative, reaction is feasible )

3.3 EXTENT OF REACTION (ε) 3.3.1. Direct chlorination Reaction: CH2CH2 (A) + Cl2 (B) → ClCH2CH2Cl (C) P = 2 atm, T = 50˚C, K2 = 1.7765 𝑦 =

;

𝑦 =

;

𝑦 =

𝐾 = ______________________________________________________________________________________ 12

Chapter 3 Thermodynamic feasibility ____________________________________________________________________________________

𝐾 = 𝐾 𝑃

=

ε = 0.9983 3.3.2. Pyrolysis Reaction: 2ClCH2CH2Cl (A) → 2CH2CHCl(B) + 2HCl (C) P = 4 atm, T = 500˚C, K2 = 1015 𝑦 =

;

𝑦 =

;

𝑦 =

𝐾 = 𝐾 𝑃 = 405300 𝜀 = 0.9999

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Chapter 3 Thermodynamic feasibility ____________________________________________________________________________________

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Chapter 4 Material balance ____________________________________________________________________________________

4.1 FLOWSHEET Process flow sheet for direct chlorination method is shown in Figure 4.1.

Figure 4.1 Process flow sheet

4.2 MATERIAL BALANCE Basis: 1000 tons/day of vinyl chloride monomer 49666.97 kg/hr of vinyl chloride monomer Percentage yield = 99% C2H4 + Cl2 C2H4Cl2

C2H4Cl2 C2H3Cl + HCl

4.2.1 Chlorination Reactor

Material balance for inlet and outlet stream of chlorination reactor is shown in Table 4.1. Table 4.1 Material balance for chlorinator reactor

Input (kg/h)

Reacted

Output (kg/h)

18855.22

18855.22

0

48051.708

47811.45

240.258(off gas)

EDC

0

0

66666.67

Total

66906.93

Ethylene Chlorine

66666.67

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Chapter 4 Material balance ____________________________________________________________________________________

4.2.2 Pyrolysis furnace Percentage conversion per pass

=

50-60%

Assumed percentage conversion

=

55%

Total EDC entering furnace

=

(66666.67 + X) kg/hr

EDC decomposed

=

0.55×(66666.67 + X) kg/hr

C2H3Cl (VCM) obtained

=

0.55×(66666.67 + X)×(62.5/99) kg/hr

Let recycled EDC be X kg/hr

Final product is 41666.7 kg/hr of VCM Assuming 1% process loss (VCM) produced

=

42016 kg/hr

0.347222(66666.67+X)

=

42016.34 kg/hr

X(recycled EDC)

=

54340.3892 kg/hr = 535.3535354 kmol/hr

Total EDC entering the furnace is 121007.059 kg/hr Unconverted EDC is 54453.177 kg/hr Material balance for inlet and outlet stream of pyrolysis furnace is shown in Table 4.2. Table 4.2 Material balance for pyrolysis furnace

Input (kg/h)

Reacted

Output (kg/h)

VCM

0

0

42016.34

HCl

0

0

24537.34

EDC Recycled EDC

66666.67

12213.4934

54453.17

Total

121007.06

54340.38 121007.06

4.2.3 Quencher EDC sprayed in the quencher is taken of the same amount as of the product from furnace. EDC used for quenching = 120000 kg/hr = 1212.12 kmol/hr Material balance for inlet and outlet stream in quencher is shown in Table 4.3. Outlet stream contains components from inlet stream and recycled stream of EDC still. Table 4.3 Material balance for quencher

Input(kg/h)

Output(kg/h)

VCM

42016.34

175460

HCl

24537.34

42016.34

EDC

54453.18

24537.34

EDC (for quenching)

121007.1

TOTAL

242013.7

242013.7

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Chapter 4 Material balance ____________________________________________________________________________________

4.2.4 HCL column Material balance for inlet and outlet stream in HCl column, where major part of HCl is removed is shown in Table 4.5. Mole fractions entering and leaving HCl column is shown in Table 4.4. Table 4.4 Mass fractions of compounds in HCl still

Feed 0.101 0.173 0.725

HCL VCM EDC

Distillate 0.978 0.008 0.013

Residue 0 0.192 0.808

D, XD C2H3Cl (0.008) C2H4Cl2 (0.013) HCl (0.978)

F, XF C2H3Cl (0.173) C2H4Cl2 (0.725) HCl (0.101)

C2H3Cl (0.192) C2H4Cl2 (0.808) HCl (0)

W, XW

Figure 4.2 Material balance in HCl still

Assuming 99% removal of HCl in the distillation column Table 4.5 Material balance for HCl still

Input F (kg/hr) 175740 42016.34 24576.66 242333

EDC VCM HCl Total

Distillate D Residue W (kg/hr) (kg/hr) 336.66 175403.349 212.54 41803.79 24576.71 0.0073 25125.91 217207.14 Toal (D+W ) = 242333

4.2.5 VCM COLUMN Material balance for inlet and outlet stream in VCM column, where major part of VCM is removed is shown in Table 4.7. Mole fractions entering and leaving VCM column is shown in Table 4.6. Table 4.6 Mass fractions of compounds in VCM still

VCM EDC

Feed 0.193 0.807

Distillate 0.968 0.032

Residue 0.004 0.996

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Chapter 4 Material balance ____________________________________________________________________________________ D, XD C2H3Cl (0.968) C2H4Cl2 (0.032)

F, XF C2H3Cl (0.193) C2H4Cl2 (0.807)

W, XW

C2H3Cl (0.004) C2H4Cl2 (0.996)

Figure 4.3 Material balance in VCM still

Feed entering VCM column is from the bottom of HCL column, neglecting small amounts of HCl. Assuming 99% removal of VCM in the distillation column. Table 4.7 Material balance for VCM still

Input F (kg/hr) 175403.34 41803.8 217207.14

EDC VCM TOTAL

Distillate D Residue W (kg/hr) (kg/hr) 1314.522 174088.82 41078.81 724.99 42393.33 174813.82 Total(D+W)= 217207.149

4.2.6 EDC column Material balance for inlet and outlet stream in VCM column, where major part of VCM is removed is shown in Table 4.9. Mole fractions entering and leaving VCM column is shown in Table 4.8. Table 4.8 Mass fractions of compounds in EDC still

EDC VCM

Feed 0.996 0.004

Distillate 0.997 0.003

Residue 0.744 0.256

D, XD C2H3Cl (0.003) C2H4Cl2 (0.997) F, XF C2H3Cl(0.256) C2H4Cl2(0.744)

C2H3Cl (0.004) C2H4Cl2((0.996)

W, XW Fig 4.4 Material balance in EDC still

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Chapter 4 Material balance ____________________________________________________________________________________ Table 4.9 Material balance for EDC still

EDC VCM TOTAL

Input F (kg/hr) 174088.83 724.98 174813.82

EDC used for quenching

= 120000 kg/hr

EDC recycled to pyrolysis

= 52999.65 kg/hr

TOTAL EDC

= 172999.7 kg/hr

Distillate D Residue W (kg/hr) (kg/hr) 173589.92 498.91 553.48 171.50 174143.40 670.41 Total(D+W)= 174813.82

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Chapter 4 Material balance ____________________________________________________________________________________

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Chapter 5 Energy balance ____________________________________________________________________________________

Energy balance of the equipment used in the process selected for the production of vinyl chloride monomer depicting the Heat of reactions and the energy at the inlet and outlet streams is mentioned in the following chapter.

5.1 PREHEATER

Figure 5.1 Preheater

Q = H = mCpΔT Q = Heat energy/hr (kJ/hr) CP = Specific heat(kJ/kg) ΔT = Temperature difference (K) Energy balance for inlet and outlet stream in preheater is shown in Table 5.1 and Table 5.2 respectively. Since the components are heated, the enthalpy of the outlet stream is more than enthalpy of the inlet stream.. INLET: Table 5.1 Energy balance of preheater (inlet)

In C 2 H4 Cl2

Mass flow (kg/h) 18855.22 48051.708

Cp(kJ/kg) 1.485 0.475

Temp (K) Hin (kJ/h)= mCpΔT 25 700000.0425 25 570614.0325 Total energy in = 1270614.075

OUTLET: Table 5.2 Energy balance of preheater (outlet)

Out C 2 H4 Cl2

Mass flow (kg/h) 18855.22 48051.708

Preheater duty

Cp (kJ/kg) 1.55 0.48

Temp (K) 60 60 Total energy out = = Total energy out - Total energy in = 1866810.57 kJ/hr = 518.55 kW

Hout (kJ/h) 1753535.46 1383889.19 3137424.65

5.2 CHLORINATOR C2 H 4

Chlorination Cl2

C2H4Cl2

Figure 5.2 Chlorinator ______________________________________________________________________________________ 21

Chapter 5 Energy balance ____________________________________________________________________________________

Energy balance for inlet and outlet stream in chlorinator is shown in Table 5.3 and Table 5.4 respectively. As the reaction is endothermic, heat contents of the outlet stream is more. The difference between energy outlet and energy inlet gives heat of reaction, neglecting heat losses. INLET: Table 5.3 Energy balance of chlorinator (inlet)

In C 2 H4 Cl2 C2H4Cl2

massflow (kg/h) 18855.22 48051.708 -

Cp (kJ/kg) 1.55 0.48 -

Temp (K) 60 60 Total energy in =

Hin (kJ/h) 1753535.46 1383889.19 0 3137424.65

OUTLET: Table 5.4 Energy balance of chlorinator (outlet)

Out C 2 H4 Cl2 C2H4Cl2

Hrxn

Mass flow (kg/h) 0 240.258 66666.67

Cp(kJ/kg) 1.55 0.48 0.758

Temp (K) 60 60 60 Total energy out =

Hout (kJ/h) 0 6919.4304 3032000.152 3038919.582

= Total energy out - Total energy in = 98505.06 kJ/h

5.3 COOLER Energy balance for inlet and outlet stream in cooler is shown in Table 5.5 and Table 5.6 respectively. INLET: Table 5.5 Energy balance of cooler (inlet)

In C 2 H4 Cl2 C2H4Cl2

Mass flow (kg/h) 18855.22 48051.708 66666.67

Cp (kJ/kg) 1.55 0.48 0.758

Temp (K) 60 60 60 Total energy in =

Hin (kJ/h) 1753535.46 1383889.19 3032000.152 3137424.65

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Chapter 5 Energy balance ____________________________________________________________________________________

OUTLET: Table 5.6 Energy balance of cooler (outlet)

Out Mass flow (kg/h) 18855.22 48051.708 66666.67

Cp (kJ/kg) Temp (K) 1.485 25 0.475 25 0.75 25 Total energy out = = Total energy out - Total energy in

C 2 H4 Cl2 C2H4Cl2 Heat removed , q

= 2520614.138 - 3137424.65 Water to be supplied, m

Hout (kJ/h) 700000.0425 570614.0325 1250000.063 2520614.138

= 616810.51 kJ/h

= q/(4.2×(60-25)) = 7342.98 kg/h

5.4 FURNACE Energy balance for inlet and outlet stream in furnace is shown in Table 5.7 and Table 5.8 respectively. Cracking of ethylene dichloride to give VCM and HCl as products i.e an exothermic reaction occurs in the furnace. INLET: Table 5.7 Energy balance of furnace (inlet)

In C2H4Cl2 C2H3Cl Cl2

mass flow (kg/h) 121007 -

Cp(kJ/kg) 0.75 -

Temp (K) 25 Total energy in =

Hin (kJ/h) 2268881.25 2268881.25

OUTLET: Table 5.8 Energy balance of furnace (outlet)

Out C2H4Cl2 C2H3Cl Cl2 Furnace duty

mass flow (kg/h) 54453.18 42016.34 24537.34

Cp(kJ/kg) 1.01 1.342 0.831

Temp (K) 500 500 500 Total energy out =

Hout (kJ/h) 27498855.9 28192964.14 10195264.77 65887084.81

= Total energy out - Total energy in = 63618203.56 kJ/h = 17671.72321 Kw

5.5 QUENCHER Energy balance for inlet and outlet stream of quencher is shown in Table 5.9 and Table 5.10 respectively. ______________________________________________________________________________________ 23

Chapter 5 Energy balance ____________________________________________________________________________________

INLET: Table 5.9 Energy balance of quencher (inlet)

In mass flow (kg/h) 175460.037 42016.34 24537.344

C2H4Cl2 C2H3Cl Cl2

Cp (kJ/kg) 1.01 1.342 0.831

Temp (K) 500 500 500 Total energy in=

Hin (kJ/h) 88607318.69 28192964.14 10195266.43 116800282.8

OUTLET: Table 5.10 Energy balance of quencher (outlet)

Out massflow(kg/h) 175460.037 42016.34 24537.344

C2H4Cl2 C2H3Cl Cl2

Cp(kJ/kg) 0.796 1.144 0.81

Temp (K) 84 84 84 Total energy out =

Hout (kJ/h) 11731959.91 4037602.209 1669520.886 17439083.01

Heat removed = Total energy out - Total energy in = -99361199.82 kJ/h

5.6 HCL STILL Energy Balance Equation: FHf + Qb = WHw+DHD+Qc F W D Hf Hw HD Qc Qb Tb

= Feed = Residue = Distillate = Enthalpy of Feed = Enthalpy of Residue = Enthalpy of Distillate = Condenser duty = Reboiler duty = Bubble point D, XD

F, XF

W, XW

______________________________________________________________________________________ 24

Chapter 5 Energy balance ____________________________________________________________________________________

Assuming R= 2.5 Q

= H = mCpΔT

FHF

= ∑ (mCpΔT) (Feed)

DHD

= ∑ (mCpΔT) (Distillate)

WHw = ∑ (mCpΔT) (Residue) Energy balance for inlet and outlet stream around HCl column is shown in Table 5.11 and the respective condenser duty calculations for the mixture at bubble point are mentioned in table 5.12 Table 5.11 Energy balance of HCl column

Feed EDC(F) VCM(F) HCL(F)

mass flow (kg/h) 175460373 42016.34 24537.344

F

242013.7213

Cp (kJ/kg) 0.796 1.144 0.81

Temp (K) 84 84 84 FHF

Hin (kJ/h) 11731959.93 4037602.209 1669520.886 17439083.03

Distillate mass flow (kg/h)

Cp (kJ/kg)

Temp (K)

Hin (kJ/h)

336.66732 212.5425 24576.71436 25125.92418

1.15 1.82 2.6

-31.32 -31.32 -31.32 DHD

-12126.08353 -12115.4326 -2001331.004 -2025572.52

EDC(D) VCM(D) HCL(D) D

Residue mass flow (kg/h)

Cp (kJ/kg)

Temp (K)

Hin (kJ/h)

175403.349 41803.79375 0.0073 217207.1501

1.64 1.458 1.9

172.3 172.3 172.3 WHw

49564075.1 10501673.2 2.389801 60065750.7

EDC(W) VCM(W) HCL(W) W

Table 5.12 Condenser duty calculations(HCl column)

Condenser 62814.81045

kg/hr

V=(R+1)D (kg/hr)

87940.73463

kg/hr

Tb(K)

241.68

K

λv at Tb (kJ/kg)

1.07E+03

kJ/kg

Qc = V×λv(Tb)

9.44E+07

kJ/hr

L=RD (kg/hr)

______________________________________________________________________________________ 25

Chapter 5 Energy balance ____________________________________________________________________________________

Reboiler : Qb

= (V+(q-1)F)λv = 1.33E+08 kJ/hr

FHF+Qb = DHD+WHW+Qc = Heat losses =

1.51E+08 kJ/hr 1.52E+08 kJ/hr 1.61E+08 kJ/hr

5.7 VCM STILL D, XD

F, XF W, XW

Assuming R= 2.5 Q

= H = mCpΔT

FHF

= ∑ (mCpΔT) (Feed)

DHD

= ∑ (mCpΔT) (Distillate)

WHw = ∑ (mCpΔT) (Residue) Energy balance for inlet and outlet stream around VCM column is shown in Table 5.13 and the respective condenser duty calculations for the mixture at bubble point are mentioned in table 5.14. Table 5.13 Energy balance of VCM column

Feed EDC(F) VCM(F) F

mass flow (kg/h) 175403.3586 41803.7975 217207.1561

Cp (kJ/kg) 1.64 1.458

Temp (K) 172.3 172.3

Hin (kJ/h) 49564077.85 10501674.1

FHF

60065751.95

Distillate

EDC(D) VCM(D) D

mass flow (kg/h)

Cp (kJ/kg)

Temp (K)

Hin (kJ/h)

1314.522 41078.8125 42393.3345

1.24 1.92

37.81 37.81

61630.57526 2982124.609

DHD

3043755.184

______________________________________________________________________________________ 26

Chapter 5 Energy balance ____________________________________________________________________________________

Residue mass flow (kg/h)

Cp (kJ/kg)

Temp (K)

Hin (kJ/h)

174088.8366 724.985 174813.8216

1.58 1.48

161.2 161.2

44339730.3 172964.021

WHw

44512694.3

EDC(W) VCM(W) W

Table 5.14 Condenser duty calculations(VCM column)

Condenser 105983.3363

kg/hr

V=(R+1)D (kg/hr)

148376.6708

kg/hr

Tb(K)

310.8

K

λv at Tb (kJ/kg)

1.93E+04

kJ/kg

Qc = V×λv(Tb)

2.86E+09

kJ/hr

L=RD (kg/hr)

Reboiler : Qb

= (V+(q-1)F)λv = 3.49E+09 kJ/hr

FHF+Qb

=

3.55E+09 kJ/hr

DHD+WHW+Qc =

2.91E+09 kJ/hr

Heat losses

6.40E+08 kJ/hr

=

5.8 EDC STILL

D, XD

F, XF

W, XW

Assuming R = 2.5 FHF

= ∑(mCpΔT) (Feed)

DHD

= ∑(mCpΔT) (Distillate)

WHw = ∑(mCpΔT) (Residue) ______________________________________________________________________________________ 27

Chapter 5 Energy balance ____________________________________________________________________________________ Table 5.15 Energy balance for EDC column

Feed EDC(F) VCM(F) F

mass flow (kg/h) 174088.83 724.985 174813.82

Cp (kJ/kg) 1.58 1.48

Temp (K) 161.2 161.2 FHF

Hin (kJ/h) 44339730.32 172964.02 44512694.35

Distillate mass flow (kg/h)

Cp (kJ/kg)

Temp (K)

Hin (kJ/h)

173589.92 553.482 174143.404

207.03 256.02

155.5 155.5 DHD

35938790.13 141702.90 36080493.04

EDC(D) VCM(D) D

Residue mass flow (kg/h)

Cp (kJ/kg)

Temp (K)

Hin (kJ/h)

498.915 171.502 670.417

1.56 4.44

425.5 425.5 WHw

331169.89 324005.36 655175.25

EDC(W) VCM(W) W

Energy balance for inlet and outlet stream around EDC column is shown in Table 5.15 and the respective condenser duty calculations for the mixture at bubble point are mentioned in table 5.16. Table 5.16 Condenser duty calculations(EDC column)

Condenser L=RD (kg/hr) V=(R+1)D (kg/hr) Tb(K) λv at Tb (kJ/kg) Qc = V×λv(Tb)

Reboiler : Qb

435358.5107 609501.9149 428.5 1.12E+04 6.82E+09

kg/hr kg/hr K kJ/kg kJ/hr

= (V+(q-1)F)λv = 1.33E+08 kJ/hr

FHF+Qb = DHD+WHW+Qc = Heat losses =

6.86E+09 6.86E+09 7.78E+06

______________________________________________________________________________________ 28

Chapter 6 Equipment design ____________________________________________________________________________________

Equipment design for the equipment in the process as per their respective flow rates and feed handling requirements are mentioned in the following chapter.

6.1 REACTOR We are considering packed bed reactor since we have gas phase heterogeneous reaction. The rate equation of the reaction is -rA = kCACB (A- Ethylene, B-Chlorine). Since we are taking Chlorine (B) in excess, rate depends on the limiting reactant, A . Rate equation is approximated to -rA = kCA . The reaction carried out in the reactor is C2H4 + Cl

50˚C , 2 atm

C2H4Cl2

The value of rate constant obtained from literature is 0.132 m3/(mol.s) . From the performance equation of the reactor (assuming plug flow behaviour) , residence time is calculated. 𝜏=𝐶

= 0.03761 ∫

∫

. .

∗ .

∗(

)

𝜏 = 40.138 seconds Catalyst volume = Volumetric flow rate of reactants×Residence time = ( 1137.99 +1113.83)× 40.138 = 25.10 m 3 Assuming void fraction in the catalytic bed to be 0.35 , Total volume of catalyst = Catalyst volume/(1- void fraction ) = 25.10 /(1-0.35) = 38.62 m3 Allowing 30% space above and below the bed for vapour, Vc= 1.3×38.62=50.213 m3 Assuming L=3d, d = 2.52 m

= 50.213

, L = 10.07 m

Length of the packed bed reactor

= 10.07 m

Diameter of the packed bed reactor

= 2.52 m

Assuming torispherical head V = 0.513 ∗ 0.06L ∗ d =0.513×0.06×10.07 ×2.522h = 1.967 m3 Total volume of the reactor = Vc + Vh = 50.213 + (2×.967)= 54.14 m3 6.1.1.Thickness calculation Shell Thickness Based on internal pressure of 2 atm Design Pressure, P = 1.1×2 = 2.2 atm = 2.2731 kgf / cm2 Thickness, 𝑡 =

+𝑐

______________________________________________________________________________________ 29

Chapter 6 Equipment design ____________________________________________________________________________________

where, j = 0.85

(joint efficiency)

f = 16 kgf / cm2

(Steel (15Cr90Mo55) IS : 1570-1961)

C = 3 mm

(corrosion factor)

2.2731 ∗ 2520 + 3 = 5.107 𝑚𝑚 (200 ∗ 16 ∗ 0.85)

𝑡=

Head Design Thickness of head, 𝑡 =

+𝐶

where, 𝑊 = Crown radius, Rc = 0.8d = 2.015 m Knuckle radius, Rk = 0.06d = 0.151 m 𝑊= 𝑡 =

. .

= 1.6628

2.2731 ∗ 2015 ∗ 1.6628 +3 200 ∗ 16 ∗ 0.85

𝑡 = 5.8 𝑚𝑚

6.2 SHELL AND TUBE HEAT EXCHANGER Tci=15˚ C

Thi=60˚C

Tho=25˚C Tco=25˚C

Figure 6.1 Shell and tube heat exchanger

Physical properties of hot and cold fluid entering the STHE are shown in Table 6.1. Table 6.1 Physical Properties of hot and cold fluid in STHE

DATA

Units

HOT FLUID

COLD FLUID

Inlet Temperaure Outlet Temperature Specific heat Density Thermal Conductivity Viscosity Mass Flow Rate Heat Flow Rate

℃ ℃ J/Kg.℃ Kg/(m^3) W/m℃ Kg/m.s Kg/s W

60 25 1239 1212 0.122 0.000601 2.315 100381.9

15 25 4314 1011 0.6034 0.001002 2.327 100381.9

______________________________________________________________________________________ 30

Chapter 6 Equipment design ____________________________________________________________________________________

Step 1: Heat Balance 𝑄 = 𝑚 𝐶 ∆𝑇 = 𝑚 𝐶 ∆𝑇 𝑘𝑔 𝐽 𝐽 𝑚 × 4314 𝑘𝑔℃ × (25℃ − 15℃) = 2.315 𝑠 × 1239 𝑘𝑔℃ × (60℃ − 15℃) 𝑚 = 2.327𝑘𝑔/𝑠 Step 2: ΔTLMTD Diagram ∆𝑇 = 𝑇 − 𝑇 = 60 − 25 = 35℃; ∆𝑇 = 𝑇 − 𝑇 = 25 − 15 = 10℃ ∆𝑇 − ∆𝑇 35℃ − 10℃ ∆𝑇 = = = 19.96℃ ∆𝑇 35℃ ln ln ∆𝑇 10℃ Step 3: ΔTLMTD Corrected 𝑇 −𝑇 60 − 25 𝑇 −𝑇 25 − 15 𝑅= = = 3.5; 𝑆 = = = 0.22 𝑇 −𝑇 25 − 15 𝑇 −𝑇 60 − 15 From 1 (Shell) – 2 (Tube) pass graph, FT = 0.8115 ∆𝑇 = ∆𝑇 × 𝐹 = 19.96℃ × 0.8115 = 16.12℃ Step 4: Allocation of Fluids Tube Side →Cold Fluid (Water) → Water Shell Side →Hot Fluid (Organic solvent) → Dichloroethane

Step 5: Find Heat Transfer Area (A) Assume U based on allocated fluids, Urange= 283.915 W/m2.oC – 851.745 W/m2.oC, therefore Uassumed = 340.698 W/m2.˚C Following, 𝑘𝑔 𝐽 2.315 𝑠 × 1239 𝑘𝑔℃ × (60℃ − 15℃) 𝑄 𝑄 = 𝑈𝐴∆𝑇 ⇒ 𝐴 = = = 18.19 𝑚 𝑈∆𝑇 340.698 𝑊 𝑚 ℃ × 16.12℃ Step 6: Tube Specification (Specifying Tube dimensions) Available Tube sizes are ¾” & 1”. Assume 1” (OD), 16 BWG. (TEMA Standards) do (OD) = ¾” = ¾ × 25.4 × 10-3 = 0.01905 m di (ID) = 0.62” = 0.62 × 25.4 × 10-3 = 0.0157 m Available Tube Length are 4 m, 4.8 m & 6 m. Assume L = 6 m Assume Triangular Pitch (Pt = 1.25 × OD) Pt = 1.25 ×¾” = 0.023812 m Step 7: Number of Tubes 𝐴 = 𝑁𝜋𝑑 𝐿 ⇒ 𝑁 =

𝐴 18.19𝑚 = = 50.68 ≈ 51 𝑡𝑢𝑏𝑒𝑠 𝜋𝑑 𝐿 𝜋 × 0.01905 𝑚 × 6 𝑚

______________________________________________________________________________________ 31

Chapter 6 Equipment design ____________________________________________________________________________________

For 1 (Shell) - 2 (Tube) pass & Pipe OD ¾”, nearest maximum number of tubes, N = 66 tubes. Shell Diameter, DS = 10”. Step 8: Heat Transfer Coefficient - Tube Side Tube Side Cross-sectional Area (per pass): 𝑁 𝜋𝑑 66 𝜋 × (0.0157 𝑚) 𝐴 = = × = 0.006𝑚 𝑁 4 2 4

Mass Velocity, 𝐺= Reynolds’ Number, 𝑁 Prandtl’s Number, 𝑁

𝑘𝑔 𝑚 2.315 𝑠 𝑘𝑔 = = 362.562 𝑚 𝑠 𝐴 0.006 𝑚

𝑘𝑔 𝑑 𝐺 0.0157𝑚 × 362.562 𝑚 𝑠 = 5695.338 = = 𝑘𝑔 𝜇 0.001002 𝑚𝑠 𝑘𝑔 𝐽 𝑚𝑠 𝐶 𝜇 4314 𝑘𝑔℃ × 0.00102 = = = 7.164 𝐾 0.6034 𝑊 𝑚℃

Heat Transfer Coefficient - Tube Side ℎ𝑑 𝑁 = = 0.023𝑁 . 𝑁 𝐾 ℎ = 1957.947 𝑊 𝑚 ℃ Step 9: Heat Transfer Coefficient - Shell Side

.

Normally, the minimum baffle spacing is ⅕th the Shell Diameter, i.e. 𝐷 10" 𝐵𝑎𝑓𝑓𝑙𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 = = = 50.8 𝑚𝑚 5 5 However, generally one can assume minimum baffle spacing, lb = 152 mm = 0.152 m 𝑡𝑢𝑏𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 6𝑚 𝑁𝑜. 𝑜𝑓 𝑏𝑎𝑓𝑓𝑙𝑒𝑠 = −1= − 1 = 38.47 ≈ 38 𝑏𝑎𝑓𝑓𝑙𝑒𝑠 𝑏𝑎𝑓𝑓𝑙𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 0.152 𝑚 Shell Diameter = 10” = 0.254 Equivalent Shell Diameter (based on triangular pitch), 4 𝐷 =

1 1 𝑑 𝑃 (𝑃 sin 60°) − 𝜋 2 2 4 1 2 𝜋𝑑

(0.01905 𝑚) 1 1 × 0.023812𝑚 × (0.023812 𝑚 × sin 60°) − × 𝜋 × 2 2 4 𝐷 = 1 2 × 𝜋 × 0.01905 𝑚 𝐷 = 0.014 𝑚 Shell Clearance [gap between adjacent tubes] (C), 4×

______________________________________________________________________________________ 32

Chapter 6 Equipment design ____________________________________________________________________________________

𝐶 = 𝑃 − 𝑑 = 0.005𝑚 Shell Side Cross-sectional Area: 𝐷 𝐶𝑙 0.254 𝑚 × 0.005 𝑚 × 0.152 𝑚 𝐴 = = = 0.00772 𝑚 𝑃 0.023812 𝑚 Mass Velocity, 𝑘𝑔 𝑚 2.315 𝑠 𝑘𝑔 𝐺= = = 297.7844 𝑚 𝑠 𝐴 0.00772 𝑚 Reynolds’ Number, 𝑘𝑔 𝐷 𝐺 0.014 𝑚 ∗ 297.7844 𝑚 𝑠 = 6964.835 𝑁 = = 𝑘𝑔 𝜇 0.000601 𝑚𝑠 Prandtl’s Number, 𝑘𝑔 𝐽 𝑚𝑠 𝐶 𝜇 4314 𝑘𝑔℃ × 0.000601 𝑁 = = = 6.104 𝐾 0.122 𝑊 𝑚℃ Nusselt’s Number, 𝑁 Step 10: Dirt Factor

ℎ 𝐷 = 0.36𝑁 . 𝑁 𝐾 ℎ = 742.216 𝑊 𝑚 ℃

=

.

Based on Organic Solvent in Shell Side & Water in Tube Side, Dirt Factor, RD = 0.003 h.ft2.oF/Btu = 0.003 × 0.1761 m2.oC/W = 0.000528 m2.oC/W Step 11: Overall Heat Transfer Co-efficient 1 1 1 1 1 = + +𝑅 = + + 0.000528 𝑚 ℃ 𝑊 𝑊 𝑊 𝑈 ℎ ℎ 1957.947 𝑚 ℃ 742.216 𝑚 ℃ 𝑊 𝑈 = 419.05 𝑚 ℃ Since, Ucalculated (419.05 W/m2.oC)>Uassumed (340.698 W/m2.oC), design is approved. Step 12: Pressure Drop – Tube Side From NRe Vs jf chart, jf = 0.011 ∆𝑃 = 𝑁 8𝑗 ∆𝑃 = 2 8 × 0.011 ×

𝐿 𝑢 + 2.5 𝜌 𝑑 2

(0.3586 𝑚⁄𝑠) 6𝑚 𝑘𝑔 + 2.5 × 1011 × 𝑚 0.0157 𝑚 2 ∆𝑃 = 4.687 𝑘𝑃𝑎

Step 13: Pressure Drop – Shell Side Assuming Baffle Cut = 25%, from NRe Vs jf chart, jf = 0.0119 𝐷 𝐿 𝑢 ∆𝑃 = 𝑁 8𝑗 𝜌 𝐷 𝑙 2 ______________________________________________________________________________________ 33

Chapter 6 Equipment design ____________________________________________________________________________________

∆𝑃 = 1 × 8 × 0.0119 ×

(0.247 𝑚⁄𝑠) 0. .254 𝑚 6𝑚 𝑘𝑔 × × 1212 × 𝑚 0.0184 𝑚 0.014 𝑚 2 ∆𝑃 = 3𝑘𝑃𝑎

6.3 CONDENSER

T=53.06˚C

T=37.81˚C

Figure 6.2 Shell and tube condenser

Physical properties of hot and cold fluid entering the condenser are shown in Table 6.2. Table 6.2 Physical Properties of hot and cold fluid in condenser

DATA Inlet Temperaure Outlet Temperature Specific heat Density Thermal Conductivity Viscosity Latent heat of vaporisation Mass Flow Rate Heat Flow Rate

Units ℃ ℃ J/Kg.℃ Kg/(m^3) W/m℃ Kg/m.s J/Kg Kg/s W

VAPOUR 53.06 37.81 551 0.01326 0.00001059 310800 1.471990781 457494.7348

WATER 20 30 4180 993 0.62 0.000734 10.94485012 457494.7348

Step 1: Heat Balance 𝑄 = 𝑚 𝐶 ∆𝑇 = 𝑚 𝜆 𝐽 𝑚 × 4180 𝑘𝑔℃ × (30℃ − 20℃) 𝑘𝑔 𝐽 = 1.472 𝑠 × 310800 𝑘𝑔 × (53.03℃ − 37.81℃) 𝑚 = 10.945𝑘𝑔/𝑠 Step 2: ΔTLMTD ∆𝑇 = 𝑇 − 𝑇 = 53.03 − 37.81 = 15.22℃; ∆𝑇 = 𝑇 − 𝑇 = 30 − 20 = 10℃ ∆𝑇 − ∆𝑇 15.22℃ − 10℃ ∆𝑇 = = = 20.322℃ ∆𝑇 15.22℃ ln ln ∆𝑇 10℃

Step 3: ΔTLMTD Corrected 𝑇 −𝑇 53.03 − 40 𝑇 −𝑇 37.81 − 20 𝑅= = = 1.525; 𝑆 = = = 0.302 𝑇 −𝑇 30 − 20 𝑇 −𝑇 53.03 − 20 ______________________________________________________________________________________ 34

Chapter 6 Equipment design ____________________________________________________________________________________

Normally, 1 (Shell) : 1 (Tube) pass is used, where LMTD correction factor (FT) can be ignored. ∆𝑇 = ∆𝑇 × 𝐹 = 20.322℃ × 0.302 = 19℃ Step 4: Find Heat Transfer Area (A) Assume U based on allocated fluids, Urange = 113.57 W/m2.oC – 283.915W/m2.oC, therefore Uassumed = 283.915W/m2.oC Following, 𝐽 𝑄 457495 𝑠 𝑄 = 𝑈𝐴∆𝑇 ⇒ 𝐴 = = = 84.82 𝑚 𝑈∆𝑇 283.915 𝑊 𝑚 ℃ × 19℃ Step 6: Tube Specification (Specifying Tube dimensions) Available Tube sizes are ¾” & 1”. Assume 1” (OD), 16 BWG. (TEMA Standards) do (OD) = 0.01905 m di (ID) = 0.01313 m Available Tube Length are 4 m, 4.8 m & 6 m. Assume L = 5 m Assume Triangular Pitch (Pt = 1.25 × OD) Pt = 1.25 ×0.01905 = 0.023812 m

Step 7: Number of Tubes 𝐴 84.82 𝑚 = = 283.45 ≈ 284 𝑡𝑢𝑏𝑒𝑠 𝜋𝑑 𝐿 𝜋 × 0.01905 𝑚 × 5 𝑚 For 1 (Shell) - 2 (Tube) pass & Pipe OD 1”, nearest maximum number of tubes, N = 346 tubes. 𝐴 = 𝑁𝜋𝑑 𝐿 ⇒ 𝑁 =

Step 8: Heat Transfer Coefficient - Tube Side Tube Side Cross-sectional Area (per pass): 𝑁 𝜋𝑑 346 𝜋 × (0.0131𝑚) 𝐴 = = × = 0.023𝑚 𝑁 4 2 4 Mass Velocity, 𝑘𝑔 𝑚 10.945 𝑠 𝑘𝑔 𝐺= = = 473.231 𝑚 𝑠 𝐴 0.023 𝑚 Reynolds’ Number, 𝑘𝑔 𝑑 𝐺 0.0131 𝑚 × 473.231 𝑚 𝑠 = 8414 𝑁 = = 𝑘𝑔 𝜇 0.000734 𝑚𝑠 Nusselt’s Number, ℎ𝑑 𝑁 = = 0.023𝑁 . 𝑁 . 𝐾 ℎ = 2859.51 𝑊 𝑚 ℃ Step 9: Heat Transfer Coefficient - Shell Side Normally, the minimum baffle spacing is ⅕th the Shell Diameter, i.e.

______________________________________________________________________________________ 35

Chapter 6 Equipment design ____________________________________________________________________________________

𝐷 0.133𝑚 = = 26.6 𝑚𝑚 5 5 However, generally one can assume minimum baffle spacing, lb = 200mm 𝑡𝑢𝑏𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 5𝑚 𝑁𝑜. 𝑜𝑓 𝑏𝑎𝑓𝑓𝑙𝑒𝑠 = −1= − 1 = 24 𝑏𝑎𝑓𝑓𝑙𝑒𝑠 𝑏𝑎𝑓𝑓𝑙𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 0.2 𝑚 𝐵𝑎𝑓𝑓𝑙𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 =

Equivalent Shell Diameter (based on triangular pitch), 1 1 𝑑 4 2 𝑃 (𝑃 sin 60°) − 2 𝜋 4 𝐷 = 1 𝜋𝑑 2 (0.01905 𝑚) 1 1 4 × 2 × 0.023812 𝑚 × (0.023812 𝑚 × sin 60°) − 2 × 𝜋 × 4 𝐷 = 1 2 × 𝜋 × 0.01905 𝑚 𝐷 = 0.014 𝑚 Shell Clearance [gap between adjacent tubes] (C), 𝐶 = 𝑃 − 𝑑 = 0.005 𝑚 Shell Side Cross-sectional Area: 𝐷 𝐶𝑙 0.1335 𝑚 × 0.005 𝑚 × 0.2 𝑚 𝐴 = = = 0.00533 𝑚 𝑃 0.023812 𝑚 Mass Velocity, 𝑘𝑔 𝑚 1.475 𝑠 𝑘𝑔 𝐺= = = 275.97 𝑚 𝑠 𝐴 0.00533 𝑚 Reynolds’ Number, 𝑘𝑔 𝐷 𝐺 0.014 𝑚 × 275.97 𝑚 𝑠 = 363858 𝑁 = = 𝑘𝑔 𝜇 0.00001059 𝑚𝑠 Nusselt’s Number, ℎ 𝐷 𝑁 = = 0.36𝑁 . 𝑁 . 𝐾 ℎ = 210.135 𝑊 𝑚 ℃

Step 10: Dirt Factor

Based on Low-Boiling Hydrocarbon in Shell Side & Water in Tube Side, Dirt Factor, RD = 0.003 h.ft2.oF/Btu = 0.003 × 0.1761 m2.oC/W = 0.000528 m2.oC/W Step 11: Overall Heat Transfer Co-efficient 1 1 1 1 = + +𝑅 = 𝑈 ℎ ℎ 2859.51 𝑊

𝑈

+ 𝑚 ℃

1 210.135 𝑊

= 289.409 𝑊

𝑚 ℃

+ 0.000528 𝑚 ℃ 𝑊

𝑚 ℃

______________________________________________________________________________________ 36

Chapter 6 Equipment design ____________________________________________________________________________________

Since, Ucalculated (289.409 W/m2.oC)> Uassumed (283.915 W/m2.oC), design is approved. Step 12: Pressure Drop – Tube Side From NRe Vs jf chart, jf = 0.013 𝐿 𝑢 ∆𝑃 = 𝑁 8𝑗 + 2.5 𝜌 𝑑 2 (0.477 𝑚⁄𝑠) 5𝑚 𝑘𝑔 ∆𝑃 = 2 8 × 0.013 × + 2.5 × 993 × 𝑚 0.01313 𝑚 2 ∆𝑃 = 9.55 𝑘𝑃𝑎 Step 13: Pressure Drop – Shell Side Assuming Baffle Cut = 25%, from NRe Vs jf chart, jf = 0.012 𝐷 𝐿 𝑢 ∆𝑃 = 𝑁 8𝑗 𝜌 = 0.88 𝑘𝑃𝑎 𝐷 𝑙 2

6.4 DISTILLATION COLUMN Relative volatility,

α= Vapour Pressure of C2H3Cl at 172.3 ℃

𝑃 = 6498.15 mmHg Vapour Pressure of C2H4Cl2 at 172.3 ℃

𝑃

= 52105 mmHg

Therefore, α= 8.01 y×

=

𝛼𝑥 1 + (𝛼 − 1)𝑥

______________________________________________________________________________________ 37

Chapter 6 Equipment design ____________________________________________________________________________________

x-y graph from equilibrium data of vinyl chloride

Equilibrium data for VCM at the operating conditions is shown in Table 6.3. Table 6.3 Equilibrium data for vinyl chloride monomer

x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Y 0 0.472 0.697 0.817 0.884 0.927 0.953 0.972 0.984 0.993 1

XD= 0.98 Xf = 0.274 XW= 0.006 XD/R+1= 0.28 From graph: Feed plate location = 5 no. of stages= 12 no. of plates= 11 Assuming 65% efficiency ______________________________________________________________________________________ 38

Chapter 6 Equipment design ____________________________________________________________________________________

Actual no. of stages= 0.65×12 = 19 Feed plate location= 0.65×5 = 8 No. of trays = 19-1 = 18 6.4.1 Estimation of Diameter ENRICHING SECTION L

: Liquid flow

G

: Vapour flow

X

: Mole fraction

Mavg

: Average molecular weight

Csb

: Flooding parameter

Unf

: Flooding velocity

𝜎

: Surface Tension

P

: parachor

Tb

: Bubble point

ꝭL

= ꝭvcmXvcm+ꝭedcXedc

ꝭvcm

= 963.63 kg/m3

ꝭedc

= 1217.61 kg/m3

Xvcm

= 0.98

Xedc

= 0.012

ꝭL

=1212.58 kg/m3

ꝭv

= PMavg/RT

Mavg

= MvcmXvcm+MedcXedc

Mvcm

= 62.5

Medc

= 99

Mavg

= 63.22

P

= 111457.5 Pa

R

= 8.314 J/mol k

Tb

= -12 ˚C

ꝭv

= 3.25 kg/m3

FLV

= (L/G)×(ꝭv/ꝭL)^0.5

R

= L/D

L

= R×D =1676.35 kmol/hr

______________________________________________________________________________________ 39

Chapter 6 Equipment design ____________________________________________________________________________________

G

= 2346.89 kmol/hr

FLV

= 0.037

Csb

= Unf×

𝜎

= P (ꝭL-ꝭv)

.

×

ꝭ ꝭ

ꝭ

(Ref. Perry Hand book edition 8th Pg.2-514 Table.2-360)

P

= 100.23

ꝭL

= 0.019gmol/cm3

ꝭv

=5.14E-05 gmol/cm3

𝜎

= 1.92

𝜎

= 13.51 dyne/cm

(Ref. Perry Hand book edition 8th Pg.14-38 Fig.14-31)

Csb

= 0.22

Therefore, Unf

= 3.92 ft/s 1.2 m/s

Operating velocity = 80%Unf Operating velocity = 0.96 m/s Flow area: Area

= G/operating velocity

G

= 12.7 m3/s

Area

= 13.26 m2

Di

=

(Area × 4/π) = 4.11 m

STRIPPING SECTION ꝭL

=ꝭvcmXvcm+ꝭedcXedc

ꝭvcm

= 2.16 kg/m3

ꝭedc

= 1149.75 kg/m3

Xvcm

= 0.006

Xedc

= 0.993

ꝭL

= 1142.23 kg/m3

ꝭv

=PMavg/RT

Mavg

=MvcmXvcm+MedcXedc

Mvcm = 99 ______________________________________________________________________________________ 40

Chapter 6 Equipment design ____________________________________________________________________________________

Medc

= 62.5

Mavg

= 62.74

P

= 111457.5 Pa

R

= 8.314 J/mol k

T

= 83.47 ˚C

ꝭv

= 2.36 kg/m3

FLV

= (𝐿/G¯)×(ꝭv/ꝭL)^0.5

𝐿

= L+qF

𝐿

= 1952.14 kmol/hr

𝐿

= 𝑉 +W

𝑉

= 182.06 kmol/hr

FLV

= 0.49

Csb

= Unf×

𝜎

= P(ꝭL-ꝭv)

𝑉

.

×

ꝭ ꝭ

ꝭ

(Ref. Perry Hand book edition 8th Pg.2-514 Table.2-360)

P

= Parachor

P

=100.23

ꝭL

= 0.018 gmol/cm3

ꝭv

= 3.76E-05 gmol/cm3

𝜎 ^ 1/4 = 1.82 𝜎

= 10.99

Csb

= 0.14

(Ref. Perry Hand book edition 8th Pg.14-38 Fig.14-31)

Therefore, Unf

= 2.73 ft/s = 0.83 m/s

Operating velocity = 80%Unf Operating velocity = 0.66 m/s Flow area: Area

= 𝑉 /operating velocity

𝑉

= 1.34 m3/s

Area

= 2.01 m2

Di

=

(Area × 4/π) = 1.60 m

6.4.2 Head Calculation t

: PdiC/200fj

______________________________________________________________________________________ 41

Chapter 6 Equipment design ____________________________________________________________________________________

f

: allowable stress

c

: chief factor

P

: Design pressure (0.1× Operating pressure)

c

=1

j

= 0.85

E

= 0.0457 mm

Di

= 4110.063172 mm

P

= 1.1 atm = 1.14 kgf/cm2

f

= 16 kgf/mm2 (IS1570-1961 15Cr90Mo)

t

= 1.72 mm

tmin

= 3 mm

6.4.3 Thickness of column t

= Pdi/ (200fj-P)

P

= 1.14 kgf/mm2

Di

= 4110.06 mm

f

= 16 kgf/mm2 (IS1570-1961 15Cr90Mo)

j

= 0.85

t

= 1.71752 mm

tmin

= 3 mm

Table 6.4 Design specifications of distillation column

DATA MATERIAL HEAD THICKNESS TTHICKNESS OF COLUMN TRAY THICKNESS(mm) CORROSION ALLOWENCE(mm) NO OF TRAYS TRAY SPACING(mm) FEED PLATE LOCATION VAPOUR AND LIQUID SPACE(mm) TOTAL HEIGHT(m) TOWER DIAMETER(m) REBOILER DUTY(kW) CONDENSOR DUTY(kW)

CARBON STEEL 3 3 10 3 19 305 5 250 6.485 4.11 1.81E+05 2.86E+09

______________________________________________________________________________________ 42

Chapter 7 Cost estimation ______________________________________________________________________________________

Many engineering design projects are developed to provide sizing information from which estimates of capital and operating costs can be made. Chemical plants are built to make a profit, and an estimate of the investment required and the cost of production is needed before the profitability of a project can be assessed Cost estimation may be defined as the process of forecasting the expenses that must be incurred to manufacture a product. Cost of various equipment is shown in Table 7.1.

Table 7.1 Cost of equipment

Cost Preheater Packed bed reactor Process pump Storage tank Furnace cooler Quencher Distillation column

No. of equipment 2 8 1 4 1 8 1 3 Total

Present cost(INR) 3156058 90744400 912103 9600000 9640000 12155884 6726704 27021150 159956300.9

7.1 DIRECT COST Total purchased equipment (TPE) = 159956300.9 INR Land = 4-8% TPE Assume Land

= 8% TPC = 0.08×159956300.9 = 12796504.07 INR

Building

= 50%TPE =0.45×159956300.9 = 79978150.43 INR

Piping

= 66%TPE = 0.66×159956300.9 = 105571158.6 INR

Electrical installation

= 10-15%TPE

Assume Electrical installation

= 13%TPE = 0.13×159956300.9 = 20794319 INR

Insulation cost

= 8-9%TPE

Assume insulation cost

= 9%TPE

______________________________________________________________________________________ 43

Chapter 7 Cost estimation ______________________________________________________________________________________

= 0.09×159956300.9 = 14396067 INR Installation cost

= 30%TPC = 0.3×159956300.9 = 47986890.26 INR

Service facility and yard improvement = 40-100%TPE Assume Service facility and yard improvement = 60%TPE = 0.6×159956300.9 = 95973781 INR Total direct cost (TDC)

= 537453170.9 INR

Direct cost

= 75%fixed capital Investment

Fixed capital cost (FCI)

= 716604227.8 INR

7.2 INDIRECT COST Engineering and supervision

= 15%TDC = 0.15×537453170.9 = 80617976 INR

Construction expenses

= 10%TDC = 0.1×537453170.9 = 53745317.09 INR

Contingency

= 5-15%FCI

Assume contingency

= 10%FCI = 0.1× 716604227.8 = 71660422.78 INR

Total indirect cost (TIC)

= 206023715.5 INR

Therefore, Fixed capital investment

= TDC+TIC =743476886 INR

7.3 WORKING CAPITAL INVESTMENT Working Capital Investment

= 10-20% of FCI

Assume Working Capital Investment WCI = 15% of FCI = 0.15× 74347688 = 111521533 INR ______________________________________________________________________________________ 44

Chapter 7 Cost estimation ______________________________________________________________________________________

7.4 TOTAL CAPITAL INVESTMENT Total Capital Investment

= FCI +WCI = 854998419.3INR

7.5 MANUFACTURING COST Fixed Charges: Depreciation

= 10% of FCI + 3% of Building Cost = 98341133.76 INR

Local Taxes

= 4% of FCI = 0.04×743476886 = 29739075.45 INR

Insurance

= 0.6% of FCI = 0.006×743476886 = 44608613.18 INR

Rent

= 8-12% Value of Rented Land & Building

Assume Rent

= 10 % Land = 1279650.407 INR

Fixed Charges (FC)

= 173968472.8 INR

7.6 DIRECT PRODUCTION CHARGES Fixed Charges

= 10-20% Total Product Cost

Assuming Fixed Charges

= 15% of Total Product Cost TPC = 1159789819 INR

Total Product Cost (TPC)

= 1159789819 INR

Operating Labor

= 5% of TPC = 0.05×1159789819 = 173968472.8 INR

Raw Materials

= 10-50% TPC

Assume Raw Material

= 30% TPC = 0.3×1159789819 = 347936946 INR

Direct Supervisory & Clerical Labor = 15% Cost of Operating Labor = 26095271 INR Utilities

= 10-20% TPC

______________________________________________________________________________________ 45

Chapter 7 Cost estimation ______________________________________________________________________________________

Assume Utilities

= 20%TPC = 0.2×1159789819 = 231957963.7 INR

Maintenance & Repair is 5% of FCI = 37173844.32 INR Operating Supplies

= 15% of Cost for Maintenance and Repair = 5576076.648 INR

Laboratory Charges

= 10-20% Operating Labor

Assume Laboratory Charges

= 20 % of Operating Labor = 34793694.56 INR

Patents &Royalties is 0 to 6% TPC Assume Patents & Royalties

= 6% TPC = 0.06×1159789819 = 34793694.56 INR

Therefore, direct production cost is equal to the sum of all the above cost. TOTAL DIRECT PRODUCTION COST = 927089658 INR

7.7 PLANT OVERHEAD COST Plant Overhead Cost

= 50-70% Total expenses for Operating Labour, Supervision & Maintenance

Assuming Plant Overhead = 60% of Total expenses for Operating Labour, Supervision & Maintenance = 1.42E+08 Therefore, Manufacturing Cost

= Fixed Charges + Direct Production Cost + Plant Overhead Cost

MANUFACTURING COST = 1243400683 INR

7.8 GENERAL EXPENSES Administrative cost

= 5% TPC

Assume Administrative cost = 5%TPC = 57989491 INR Distribution & Marketing Cost = 2-20% of TPC ______________________________________________________________________________________ 46

Chapter 7 Cost estimation ______________________________________________________________________________________

Assume Distribution & Marketing Cost to be 20% of TPC = 231957963.7 INR Research & Development Cost = 2-5%TPC Assume R&D cost to be 3 % of TPC = 34793695 INR Therefore, General Expenses are equal to the sum of all the above cost. GENERAL EXPENSES, Tge = 324741149.2 INR

7.9 TOTAL PRODUCT COST Total Product Cost

= Tmc + Tge = 1568141833 INR

7.10 GROSS EARNING/ INCOME Whole sale selling price of VCM/ton = 6550 Rs Total Income = (Selling price) × (Quantity of Product Manufactured) = 1965000000 INR (Assume 300 working days) Gross Income

= (Total Income) – (Total Production Cost) = 1965000000 – 15681418 = 396858167.4 INR

Current Tax Rates

= 45%

Taxes

= 45% of Gross Income = 0.45×396858167.4 = 178586175.3 INR

Therefore, Net Profit

= Gross Income –Taxes = 218271992.1 INR

7.11 RATE OF RETURN Rate of return is a profit on an investment over a period of time, expressed as a proportion of the original investment. Rate of Return = Net profit ×

= 25.528 %

7.12 PAYOUT PERIOD The payback period is the length of time required to recover the cost of an investment. The payback period of a given investment or project is an important determinant of whether to undertake the position or project, as longer payback periods are typically not ______________________________________________________________________________________ 47

Chapter 7 Cost estimation ______________________________________________________________________________________

desirable for investment positions. Appropriate payback period for a medium production plant is 3-6 years. Calculated parameters of cost estimation in shown in Table 7.2. Payout Period

= = 3.91 years

PAYOUT PERIOD = 3.91 years

Table 7.2 All costs involved

S.NO

DESCRIPTION

PRICE (INR)

1

TOTAL DIRECT COST

537453170

2

TOTAL INDIRECT COST

206023715

3

FIXED CAPITAL INVESTMENT

716604227

4

WORKING CAPITAL COST

111521533

5

TOTAL CAPITAL INVESTMENT

854998419

6

FIXED CHARGES

173968472

7

TOTAL GENERAL EXPENSES

324741149

8

MANUFACTURE COST

1243400683

9

TOTAL PRODUCT COST

1568141833

10 11

GROSS INCOME NET PROFIT

396858167 218271992

12

PAY BACK PERIOD

3.91712382

13

RATE OF RETURNS

25.53%

______________________________________________________________________________________ 48

Chapter 8 Piping instrumentation and diagram ______________________________________________________________________________________

Instrumentation and process control engineering plays a vital role in controlling the process for desired output from the plant. Instrumentation is provided to monitor the vital process parameters during plant operations. They may be incorporated in automatic control loops or used for the manual monitoring of process parameters. They may also be a part of automatic computer data acquisition. Instruments monitoring critical process parameters will be fitted with automatic pop-ups and enunciations to alert the operator at critical and hazardous situations in prior. The main objective of the designer when specifying instrumentation and control schemes are, to keep process parameters within the operating limit, so as to detect dangerous situations that may develop due to process deviations and to provide alarms/buzzer and automatic fail safe systems. And also to maintain the product composition within the specified quality standards and to operate the plant at the lowest production cost. The main objectives of designer when specifying instrumentation and process control are, a. Monitor and control the process parameters within the operable range on real time basis. b. Maintaining the product quality within the specifications. c. Ensure optimum production cost with integrated fail safe mechanisms. Eliminating all the pit-falls that may rise due to manual intervention and properly taken care of such situations and alert by prior signalling and controlling catastrophic failures automatically before-hand.

8.1. INSTRUMENTATION 8.1.1 Flow measuring instruments For measuring flow of combined feed to the reactor, mass flow meter will be used which is sensitive in measuring and the measured readings will be fed back to the control system. Flow measuring instruments are available in various categories like variable head flow meter, electromagnetic flow meters, etc and in various ranges; in this process stainless casing of the variable head flow meter is suitable for good results. 8.1.2 Temperature measurement For measuring the temperature of flow streams in any equipment, resistance thermometers or Thermocouples are used. Platinum- platinum rhodium thermocouples are used in most industries, since thermocouples give excellent transmission accuracies than other temperature measuring instruments. For measuring the temperature of the reactor the same thermocouples are used. ______________________________________________________________________________________ 49

Chapter 8 Piping instrumentation and diagram ______________________________________________________________________________________

8.1.3. Pressure Measurement The bourdon tube measuring elements are used in combination with the differential pressure transmitters with stainless steel internal components are used for the measuring the reactor bed pressure drops in mm water column. Also electronic pressure transmitters are also used for transferring the pressure reading to the panel at high accuracy. 8.1.4. Concentration measurement The composition of the outlet stream can be analysed by HPLC chromatography, which gives more accurate and fast feedback.

8.2. CONTROL SYSTEMS For the equipment used, the following controls are required for safe and efficient operation: 8.2.1. Flow controller Flow controllers are used to control the flow of the stream. Flow controllers are used to maintain specified molar ratios of the reactants. Flow controllers are used here while pumping the feed to the pyrolysis furnace to maintain the flow rate of the reactants i.e., preheated ethylene and chlorine and also in the condensers of the distillation columns to obtain the distilled products at required flowrates. 8.2.2. Temperature controller These are devices that controls the passage of heat energy into or out of the space and adjusts to the desired temperature. Air conditioners, space heaters, refrigerators, water heaters, etc. are examples of devices that control temperature. In the process temperature controllers are used while preheating to heat the reactants to desired temperature, to cool the product formed to required temperature after the reaction using a cooler, in the pyrolysis furnace so that cracking can take place at higher temperatures in the furnace and in the reboilers of the distillation columns so that desired temperature of the residues is obtained. 8.2.3. Pressure controller Pressure controllers are controllers which quickly and automatically provide a pressure based on the supply pressure. These maintain the consistency in the pressure. In the distillation columns when the distilled products are removed, a constant pressure has to be maintained in the condensers using pressure controllers.

______________________________________________________________________________________ 50

Chapter 8 Piping instrumentation and diagram ______________________________________________________________________________________

8.2.4. Ratio controller It is a controller where disturbances are measured and the ratio is held at a desired set point by controlling one of the streams. The ratio controllers are used here to control the feed ratio of ethylene and chlorine before heating so that it is sent to the reactor. 8.2.5. Level controller Level controllers are instruments which monitor fluid levels in vessels, and to controls the level as required. Level controllers are used in the packed bed reactor where level has to be maintained so that the reaction takes place. These are also used in the storage vessel where the level should be maintained so that the product formed should be stored and overflow does not take place. 8.2.6. Interlocks of the reactor Interlocks are control functions that prevent the normal operation of other control or operating functions. They are used to ensure personnel or equipment safety or to protect the operability of the process. Interlocks will often be used to prevent the pump from operating if a valve is closed on either suction or the discharge side of the pump or if the level of the tank feeding the pump is low. Interlocks do not take intermediate action, but prevent either the operator or a control function from operating, are often referred to as permissive. If interlocks are required for personnel safety, they should normally be separate from the regulatory control system in order to protect against control system failure. 8.2.7. Emergency Shutdown of the reactor system Because of the instability of the exothermic reactions and the possibility of a runaway reaction, the reactor is often equipped with an emergency shutdown system, as well as other safety equipment, such as burst diaphragms and release containment systems. The purpose of the emergency shutdown system is to stop the reaction in the event of a runaway. The system may accomplish this by quickly reducing the temperature, injecting some material into the reactor, which will reduce the rate of the reaction; or venting the reactor to reduce the pressure. The emergency shutdown system must be highly reliable. In almost all industries computer based distribution control systems and programmable logic controllers with human interface systems are installed for easy control through computers. In this system, the hardware requirements for pneumatic, electronic, and microprocessor-based controllers are employed, all pneumatic signals in the range of 3 to

______________________________________________________________________________________ 51

Chapter 8 Piping instrumentation and diagram ______________________________________________________________________________________

15 psig , the energy needed to operate these pneumatic components is a source of clean, dry air at a pressure of about 20 psig. 8.2.8 Alarms An extensive amount of the software in modern controllers is devoted to detecting and reporting a problem in the form of an alarm. The alarm takes the form of a visual signal, an audible signal (beeping horn), or the auction of a switch.

8.3. DATA STORAGE Long term storage of the transients can be obtained easily with a digital computer; this task is referred to as archiving. The automatic storage of critical process control variables on disk or tape can be retrieved later to explain process operating difficulties. The computer can also be used to automatically record or log the type and location of an alarm, the time of a process alarm, the time of acknowledgement of an alarm, and the time it was cleared by operator intervention.

______________________________________________________________________________________ 52

Chapter 9 Plant location and layout ______________________________________________________________________________________

We have chosen Ratnagiri as our location for manufacture of vinyl chloride monomer. It is located in the state of MAHARASHTRA in INDIA shown in Figure 9.1.

Figure. 9.1 Plant location

The reasons for selecting this location are listed below. 1) Geography and climate: Ratnagiri is a port city situated on the Arabian Sea coast in the southwestern part of Maharashtra. It has an average elevation of 11 meters. 2) Transport - It has good railways and road connectivity. It is 330 Km north of Mumbai Connected with road and rail links with JNPT Mumbai Vadodara. 3) Availability of labour - Local labour is easily available as it has many chemical industries. 4) Present Industrial condition – Ratnagiri is an industrial area. Some of the renowned industries names include JSW Group, Ultratech Cement, Finolex Group, VAV Life Sciences, Georg Fischer, Gadre Marine. A nuclear project is also undertaken there named Jaitapur Nuclear Power Project. 5) Thermal power supply – Ratnagiri Gas and Power Private Limited (RGPPL) is a joint venture of NTPC Limited, GAIL and Government of Maharashtra. RGPPL owns India’s largest gas based power plant and the LNG Gasification terminal at Dabhol. It’s three power blocks supply a combined of 1967 MW of electricity to India’s western grid.

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Chapter 9 Plant location and layout ______________________________________________________________________________________

9.1 PLANT LAYOUT Plant layout can be defined as the physical location or configuration of departments, workstations and equipment in the production process. It is well defined arrangements of physical resources which are used to produce the product. Plant layout is placing the right equipment, coupled with the right method, in the right place to permit the processing of product in the most effective manner through the shortest possible distance and in the shortest possible time. Factors considered for plant layout are: 

Organization



Location of departments.



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



Convenience of operation and maintaince



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



Health and safety considerations



Future expansion

9.2 PLANT LAYOUT PROCEDURE A number of procedures have been developed to facilitate the planning (Design) of plant layouts. Following steps are generally considered in designing a plant layout. The type of data and information required depends upon the nature of the layout problems. In case planning a new plant layout, the data required will be very elaborate and will cover the manufacturing particulars along with growth rate required. If the layout problem is for making minor alterations in the layout, the data required will be limited to manufacturing methods, sequence of operations, equipment involved etc., Product analysis consists of breaking down the product into sub-assemblies and the subassemblies into their individual parts. Product analysis determines the operations necessary for the production of each part. Production process is then analyzed. Since the purpose of the flow pattern is to plan the movement of the raw material and product parts in a straight line through the plant. Material flow pattern should be designed to ensure minimum movement in terms of distance and time. Inter relationship between the work areas is also planned while designing the individual operations or work areas. ______________________________________________________________________________________ 54

Chapter 9 Plant location and layout ______________________________________________________________________________________

Similarly, the inter relationship between related groups of operations should be worked out. At this stage, flow diagram also should be co coordinated with the material flow pattern. After installing the layout there may be possibility that it may not work as planned, people may not follow the method they are supposed to, material does not arrive as planned. it may result in delays. In such circumstances management should try to find out the reasons for such delays, the activities which have to be taken care off and evaluate the consequences. Action should be taken to rectify the mistakes, to minimize their effect on the execution of the plant. Sample of the plant layout is shown in Figure 9.2.

Fig 9.2 PLANT LAYOUT

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Chapter 9 Plant location and layout ______________________________________________________________________________________

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Chapter 10 Health and safety ______________________________________________________________________________________

Vinyl chloride is an OSHA- regulated material. Current OSHA regulations require that no employed be exposed to vinyl chloride concentrations greater than 1 ppm over any 8 hours, or 5ppm averaged over any period not exceeding 15 min. monitoring is required at all facilities where vinyl chloride is produced or PVC is processed. The monitoring may be discontinued only when at least two when at least two consecutive determinations made not less than five working days apart, show exposure at or below the action level of 0.5ppm contact with liquid vinyl chloride is prohibited. Chronic exposure to vinyl chloride produced Raynaud’s syndrome, lysis of the distal bones of the fingers and a fibrosing dermatitis. However, these effects are probably related to continuous intimate contact with skin. Chronic exposure is reported to have produced a rare cancer of the liver in a small number of workers after continuous exposure for many years to large amount of vinyl chloride gas. Exposure to vinyl chloride can be readily reduced by feasible engineering controls of poor practice to below the OSHA acceptable levels. Use of closed systems or laboratory goods that have protection factors adequate to prevent worker exposure are recommended whenever exposure is above the permissible OSGA limit and cannot be reduced by feasible engineering practices, respirators are required and must be used in accordance with a standard respirator program. The danger comes during the production process. Three basic steps are necessary to the production of vinyl chloride products. First, the vinyl chloride, or monomer, is produced in a closed process. The vinyl chloride monomer is then shipped as a compressed liquefied gas to plants/locations that produce the polyvinyl chloride resin. Second, batches of vinyl chloride are polymerized by mixing them with catalysts in giant vats or reactors. After drying, the polyvinyl chloride resins are compounded by the addition of stabilizers, lubricants, and plasticizers. Third, the polyvinyl chloride resins are fabricated into a myriad of finished products. This step involves a large number of industrial processes. Vinyl chloride is flammable when expected to heat, flame or oxidizing agents. Large fires of the compared are very difficult to extinguish. Vapors represent a severe explosion hazard.. Because of possible peroxides formation. Vinyl chloride should be transported or handled under an inert atmosphere .However, stabilizer can be added to prevent polymerization. Because of the world side production and transport of vinyl chloride the higher temperature of certain climates may require that small amount of phenols or other stabilizer be added to the vinyl chloride.

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Chapter 10 Health and safety ______________________________________________________________________________________

10.1 Controlling the Hazard The best method of controlling exposure to vinyl chloride is achieved through the substitution of a less hazardous material. Where this is not possible, engineering controls such as enclosure of the source of exposure and local exhaust ventilation should be implemented. As a last resort, employers should introduce administrative controls and provide personal protective equipment such as non-porous gloves, goggles, non-porous aprons, sleeves, boots, and, where vinyl chloride levels cannot meet the OSHA Standard, respiratory protection. 10.2 The OSHA Standard The OSHA Standard applies to the manufacture, reaction, packaging, re-packaging, storage, handling, or use of vinyl chloride in such operations as molding, extrusion, mixing, and calendaring. The Standard does not apply to the handling or use of fabricated products made of polyvinyl chloride such as thermos forming or blister packaging.

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References ______________________________________________________________________________________

REFERENCES 1. Nomenclature of Organic Chemistry : IUPAC Recommendations and Preferred Names 2013 (Blue Book). Cambridge: The Royal Society of Chemistry. 2014. pp. 649, 717. doi:10.1039/9781849733069-FP001. ISBN 978-0-85404-182-4. For example, the name ‘ethanolamine’, which is still widely used, is badly constructed because of the presence of two suffixes; it is not an alternative to the preferred IUPAC name, ‘2-aminoethan-1-ol’. 2. "Ethanolamine MSDS" (PDF). Acros Organics. Archived from the original (PDF) on 2011-07-15. 3.

Hall, H.K., J. Am. Chem. Soc., 1957, 79, 5441.

4.

R. E. Reitmeier; V. Sivertz; H. V. Tartar (1940). "Some Properties of Monoethanolamine and its Aqueous Solutions". Journal of the American Chemical Society. 62 (8): 1943–1944. doi:10.1021/ja01865a009.

5.

Sigma-Aldrich Co., Ethanolamine. Retrieved on 2018-05-24.

6.

"NIOSH Pocket Guide to Chemical Hazards #0256". National Institute for Occupational Safety and Health (NIOSH).

7. “Ethanolamine". Immediately Dangerous to Life and Health Concentrations (IDLH). National Institute for Occupational Safety and Health (NIOSH). 8. Cough, Cold, and Allergy Preparation Toxicity at eMedicine 9. Klaus Weissermel; Hans-Jürgen Arpe; Charlet R. Lindley; Stephen Hawkins (2003). "Chap. 7. Oxidation Products of Ethylene". Industrial Organic Chemistry. Wiley-VCH. pp. 159–161. ISBN 3-527-30578-5. 10. http://lipidlibrary.aocs.org/Lipids/pe/index.htm 11. Calignano, A; La Rana, G; Piomelli, D (2001). "Antinociceptive activity of the endogenous fatty acid amide, palmitylethanolamide". European Journal of Pharmacology. 419 (2–3): 191–8. doi:10.1016/S0014-2999(01)009888. PMID 11426841. 12. "Ethanolamine". Occupational Safety & Health Administration. 13. Carrasco, F. (2009). "Ingredientes Cosméticos". Diccionario de Ingredientes Cosméticos 4ª Ed. www.imagenpersonal.net. p. 306. ISBN 978-84-613-4979-1. 14. http://www.inclusive-science-engineering.com/process-technology-to-produceethanolamines-by-reaction-of-ammonia-and-ethylene-oxide 15. https://www.cdc.gov/niosh/npg/npgd0256.html 16. Peters, M.S., Timmerhaus, K.D., West, R.E., Timmerhaus, K. and West, R., 1968. Plant design and economics for chemical engineers (Vol. 4). New York: McGraw-Hill. 17. Kirk-Othmer Encyclopaedia of Chemical Technology Volume 4 ______________________________________________________________________________________ 59

References ______________________________________________________________________________________

18. Dryden's Outlines of Chemical Technology for the 21st Century, M. Gopala Rao, Marshall Sittig 19. Groggins, P.H. (ed.), “Unit Processes in Organic Synthesis”, 5th edition 20. R.H. Perry and D.W. Green, “Perry’s Chemical Engineering Handbook”, 5th edition, McGraw Hill 1997. 21. Narayanan, K.V., 2004. A textbook of chemical engineering thermodynamics. PHI Learning Pvt. Ltd.

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