trichloroethylene thesis
TRANSCRIPT
1
1. INDUSTRY PROFILE
THE DCW STORY GOES back to 1925 when the foundation stone of India's first Soda
Ash factory at Dhrangadhra - a small principality in Gujarat in West India - was laid. The
plant was taken over in 1939 and run under the name Dhrangadhra Chemical Works. To
wish the venture luck, the company adopted the horse shoe as their corporate logo, which
stands till today, and is widely recognized as a symbol of excellence.
It is said that every oak begins as an acorn. The process of growth received a major
impetus in 1959 with the commissioning of the chlor-alkali plant at Sahupuram in the
southern state of Tamilnadu. At that time, the area was completely barren. Today this
complex has made its mark on the chemical map of India as well as the world.
Growth at the chlor-alkali complex was rapid as between 1965 and 1970 three plants
were erected that turned the co-product chlorine into a money spinner; a liquid chlorine
plant in 1965, the country's first Trichloroethylene plant in 1968 and an integrated PVC
resin plant in 1970-making the company one of the first in the nascent petrochemicals
field.
In the same year, 1970, the company set up a plant to manufacture upgraded ilmenite,
the first of its kind in Asia, and even today, one of the few of its kind in the world. In 1986,
to reflect the expanded activity spectrum, and its emergence as a multi-product and multi-
locational company, the corporate name was changed simply to DCW Ltd.
PRODUCTS INSTALLED
CAPACITY MT PRODUCTION MT
CAUSTIC SODA DIVISION
CAUSTIC SODA LYE 100000 77612
CAUSTIC SODA SOLID 155
CAUSTIC SODA FLAKES 25910
LIQUID CHLORINE 36000 18963
TRICHLOROETHYLENE 7200 4715
UPGR & W.G.ILM 48000 36394
UTOX 600 1612
IRON OXIDE 297
PVC DIVISION 90000 85758
Table 1.1 Industry Products
2
2. INTRODUCTION
2.1. HISTORY OF TRICHLOROETHYLENE
By 1896, work in Germany led to chlorinated solvents by partial or complete chlorination
of acetylene, and in 1908 to a full-scale plant producing 1,1,2-trichloroethene. Pioneered by
Imperial Chemical Industries in Britain, its development was hailed as an anesthetic
revolution. Originally thought to possess less hepatotoxicity than chloroform, and without
the unpleasant pungency and flammability of ether, TCE use was nonetheless soon found
to have several pitfalls. These included promotion of cardiac arrhythmias, too low volatility
for quick anesthetic induction, reactions with soda lime used in carbon dioxide absorbing
systems, prolonged neurological dysfunction when used with soda lime, and evidence of
hepatotoxicity as had been found with chloroform.
The introduction of halothane in 1956 greatly diminished the use of TCE as a general
anesthetic. TCE was still used as an inhalation analgesic in childbirth given by self-
administration. Fetal toxicity and concerns for carcinogenic potential of TCE led to its
abandonment in the 1980s.
Due to concerns about its toxicity, the use of trichloroethylene in the food and
pharmaceutical industries has been banned in much of the world since the 1970s.
Legislation has forced the substitution of trichloroethylene in many processes in Europe as
the chemical was classified as a carcinogen carrying an R45 risk phrase.
Fig 2.1 Chemical Structure of TCE
3
2.2. PHYSICAL PROPERTIES
CHEMICAL FORMULA CHCI = CCI2
PHYSICAL STATE AT ROOM TEMP liquid
APPEARANCE ALMOST colorless
FLASH POINT nonflammable
VAPOR PRESSURE AT 20°C 68mmHg
BOILING POINT @760mmHg 86.7°C
FREEZING POINT -87.1°C
MOLECULAR.WT 131.4
Sp.GRAVITY @ 20°C 1.464
Sp.HEAT @ 80°C 0.225cal/gm°C
VISCOSITY @ 20°C 0.58cP
REFRACTIVE INDEX@20°C 1.4782
Table 2.2.1 Physical Properties of TCE
4
2.3. CHEMICAL PROPERTIES
TCE is highly soluble in ether and alcohol and is miscible with most organic solvents while
being practically insoluble in water
Compound (1) decomposes to form dichloroacetyl chloride, which in the presence
of water decomposes to dichloroacetic acid and hydrochloric acid (HCl) with
consequent increases in the corrosive action of the solvent on metal surfaces.
Compound (2) decomposes to yield phosgene, carbon monoxide, and hydrogen
chloride with an increase in the corrosive action on metal surfaces.
It is not readily hydrolyzed by water. Under pressure at 150◦C, it gives glycolic acid
with alkaline hydroxides.
Liquid trichloroethylene has been polymerized by irradiation with 20-keV x-rays
Fig 2.3.1 Decomposed Product
5
2.4. PRODUCTION METHODS
Prior to the early 1970s, most trichloroethylene was produced in a two-step process from
acetylene. First, acetylene was treated with chlorine using a ferric chloride catalyst at 90 °C
to produce 1,1,2,2-tetrachloroethane according to the chemical equation
HC≡CH + 2 Cl2 → Cl2CHCHCl2
The 1,1,2,2-tetrachloroethane is then dehydrochlorinated to give trichloroethylene. This can
either be accomplished with an aqueous solution of calcium hydroxide
2Cl2CHCHCl2 + Ca(OH)2 → 2ClCH=CCl2 + CaCl2 + 2 H2O
or in the vapor phase by heating it to 300-500°C on a barium chloride or calcium chloride
catalyst
Cl2CHCHCl2 → ClCH = CCl2 + HCl
Today, however, most trichloroethylene is produced from ethylene. First, ethylene is
chlorinated over a ferric chloride catalyst to produce 1,2-dichloroethane.
CH2 = CH2 + Cl2 → ClCH2CH2Cl
When heated to around 400 °C with additional chlorine, 1,2-dichloroethane is converted to
trichloroethylene
ClCH2CH2Cl + 2 Cl2 → ClCH = CCl2 + 3 HCl
This reaction can be catalyzed by a variety of substances. The most commonly used
catalyst is a mixture of potassium chloride and aluminum chloride. However, various forms
of porous carbon can also be used. This reaction produces tetrachloroethylene as a
byproduct, and depending on the amount of chlorine fed to the reaction, tetrachloroethylene
can even be the major product. Typically, trichloroethylene and tetrachloroethylene are
collected together and then separated by distillation.
6
3. CHEMICAL PROCESS
3.1. VARIOUS PROCESSES
Chlorination of Acetylene
Chlorination of Ethylene
Oxychlorination of Ethylene or Dichloroethane
3.2. CHLORINATION OF ACETYLENE
Reaction involved
Step1.Chlorination
Step2. Dehydrochlorination
The above both reactions are exothermic reactions.
C2H
2 + 2Cl
2 C
2H
2 Cl4 ΔH= -402kJ/kmol
90°c
FeCl3
2C2H
2Cl
4 + Ca(0H)
2 2C
2HCl
3 + CaCl
2 + 2H
2O ΔH= -150kJ/kmol
80-85°c
7
3.3. REASONS FOR THIS PROCESS SELECTION
Selected process - Chlorination of Acetylene
Acetylene generated from CaC2 which easily produced from lime and Coal
Acetylene purity became 99.5% v/v
Higher yield of Tetrachloroethane which is feed for TCE
Lime slurry from Acetylene generation is reused for Dehydrochlorination
Higher equilibrium conversion of 95% without catalyst
Yield of TCE is 90% with 99.6% purity
3.4. PROCESS DESCRIPTION
3.4.1. RAW MATERIALS
Acetylene gas (C2H2)
Chlorine gas (Cl2)
Catalyst (FeCl3)
Stabilizers (Phenol, Thymol)
3.4.2. PROCESS PARAMETERS
Step1. Chlorination of Acetylene
Temperature : 80 to 90°C
Feed molar ratio : Acetylene to Chlorine, 1 : 2
Pressure : 1.1 bar
Step2. Dehydrochlorination of Tetrachloroethane
Temperature : 90°C
Pressure : 1.1 bar Feed
Molar ratio (lime to Tetrachloroethane): 1 : 2
8 Fig 3.4.1 PFD for TCE
9
3.4.3. STEPS INVOLVED IN PROCESS
The acetylene-based process consists of two steps
Step1. First acetylene is chlorinated to 1, 1, 2, 2-tetrachloroethane. The reaction is
exothermic (402 kJ/mol = 96 kcal/mol but is maintained at 80–90°C by the vaporization of
solvent and product. Catalysts include ferric chloride.
Here, Acetylene gas (C2H2) is dried (<40ppm water) with diluted sulphuric acid
(80%) and sent to the bottom of synthesis column with two molar times of Chlorine gas
(Cl2). But it‟s indirectly contacted with teterachloroethane medium that in liquid phase. So,
the reaction followed by absorption of gases with tetrachloroethane. The unreacted gases
synthesized in recovery column. Product is continuously discharged from the overflow line
which attached in the middle of reactor.
This reaction is highly exothermic in nature. So, the reaction mixture continuously
cooled by the open type heat exchanger upto 85°C. Finally product tetrachloroethane stored
in closed container because of their higher volatility and presents of free chlorine.
Step2. The product is then dehydrochlorinated to trichloroethylene at 96–100°C in aqueous
bases such as Ca(OH)2. The yield of trichloroethylene is about 94% based on acetylene. A
significant disadvantage of the alkaline process is the loss of chlorine as calcium chloride
(CaCl2).
In this section aqueous lime with 18% concentration that preheated to 80°C. After
preheating another reactant tetrachloroethane was fed continuously as per the theoretical
requirement. Here Tetrachloroethane (C2H2Cl4) is the limiting reactant. So lime is added
excess quantity.The crude Trichloroethylene obtained from the reactor top in vapor phase,
with the application of total condenser. Also crude product analyzed for it acidity nature
because of the dehydrochlorination reaction takes place.
Finally, the crude Trichloroethylene is charged in the fractionating column and the
product pure Trichloroethylene separated at the top of the column with 99.5% purity.
10
3.5 APPLICATIONS
o Metal cleaning, finishing and vapor degreasing
Cold, ultrasonic and vapor degreasing of metal parts between fabrication steps and
before finishing or assembly, surface preparation of metal parts for galvanizing,
anodizing, electro plating, painting and bonding.
For quick removal of oils and waxes in dip coating of rust preventing formulations,
the oil and grease covered parts are placed in an atmosphere of TCE vapor. The hot
vapor condenses immediately on the cold metal surface, dissolving the oils and
greases and flushing them. The work heats to the vapor temperature in minutes,
leaving the work clean, dry and ready for further treatment.
o Solvent extraction
Extraction of vegetable oils, waxes, animal fats, caffeine, cocoa butter, essential
oils, pharmaceuticals, for recovery of oils and grease from waste, rags and paper,
turnings etc.
Chemical Processes
In dehydration and purification of many chemicals, as a reactant for manufacture of various
halohydrocarbons and other chemicals, as an additive for promoting polymerization of
monomers etc…
Other uses
TCE is a good heat transfer medium in the temperature range of -73°C to +120°C
and is used in simulating out door conditions in material testing.
Used as a coolant for cutting tantalum and other special metals.
Used as a freezing point depressant, especially in fire extinguisher fluids.
Used in making rubber cements, bone glue and adhesive activators.
As a solvent for paints, lacquers, dyes, oils, fats, waxes, resins, halohydrocarbons
and many other chemicals.
11
4. MATERIAL BALANCE
4.1. BATCH REACTOR
Basis: 4380 kg of
Crude TCE (3000 liters)
2(167.8) 74.1 2(131) 111 2(18)
As per the reaction
335.6 kg of Tetra required to produce 262.7 kg of TCE
So for the production of 4380 Kg of TCE = (4380*335.6)/262.7
= 5595.46 kg
(Density of Tetra = 1.58) = 3.542 m3
But in actual process, they are fed 3600 liters of Tetrachloroethane (limiting reactant) for
3000 liters of crude TCE production in one Batch
Reactor conversion = (3.542 / 3.6)*100 = 98.4%
Amount of solid CaCl2 formed = (111/262.7)*4380
= 1850.7 kg
BATCH
REACTOR
4380 Kg of crude TCE
6864 Kg of 18% lime
5595 Kg of Tetra
15% of CaCl2 6230 Kg
2C2H
2Cl
4 + Ca(0H)
2 2C
2HCl
3 + CaCl
2 + 2H
2O ΔH= -150kJ/kmol 80-85°c
Fig 4.1.1 Batch Reactor
12
4.2. DISTILLATION
Here the crude Trichloroethylene contains the following components,
1. Trichloroethylene (TCE)
2. Dichloroethylene (vinyl dine chloride)-ligherends
3. Perchloroethylene (PERC)
Basis: 21 ton per day of pure TCE,
TCE required = 21000/24
= 875 kg/hr
In Fractionator 2
F1
F
F2
B
D
Fractionator 1
Freactionator 2
Fig 4.2.1 Fractionator column
13
Top product contains 99.4% TCE
Bottom contains 3% TCE
Feed contains 93.0% TCE
7.0% Perchloroethylene
TCE balance
0.93F2 = 875 + 0.03B ---------> 1
Top product obtains
0.994D = 875
D = 880.28
Therefore PERC in Distillate = 5.28 kg/hr
PERC balance
0.07F2 = 5.28 + 0.97B ----------->2
From equation 1 and 2
F2 = 942.87 kg/hr
B = 62.59 kg/hr
Fractionator 1
We assume,
All of Perchloroethylene goes in the bottom
Feed composition
TCE = 91.74%
14
Dichloroethylene = 1.65%
PERC = 6.60%
PERC balance
0.066 F = 0.07 F2
F = 1000 kg/hr
Product (TCE) Balance
0.917 F = 0.93 F2 + 0.1 F1 ---------3
From overall balance
F1 = 401.22 kg/hr
RESULTS FROM F1 and F2
Fractionators resulting the following composition
TCE from F1 = 40.12 kg/hr
Dichloroethylene from F1 = 361.1 kg/hr
Product (TCE) from F2 top = 875 kg/hr with 99.5% purity
Heavier ends (Perchloroethylene) = 1000 – 875
= 275 kg/hr
15
5. ENERGY BALANCE
5.1. BATCH REACTOR
Energy balance across a reactor
Basis: 8 hrs- batch
Heat capacity of reactants + Heat supplied form of steam + Heat of reaction (Exothermic)
= Heat capacity of products + Heat loses in reactor
Sensible Heat = m Cp ΔT
Latent Heat = m λ
(In reactor conversion is 98% only)
Take a basis for 8 hours for the time of one batch completion
3600 liters of Tetrachloroethane is converted in 3000 liters of Trichloroethylene
The respective density of the above compounds likely 1.58 and 1.48 gm/cc
ENERGY IN INPUT
Reactants Mass kmol Cp kJ/kmolK Heat Q, kJ
Tetrachloroethane 5760 34.28 0.97 27936
Lime 18% solution 6864 244.27 103.16 2.52×106
Steam supplied 532 28.5 40626 (λ) 1.20×106
Heat of
Reaction(Exothermic)
5760 34.28 150000 98%
conversion
5042
Total energy 3.75×106
Table 5.1.1 Reactor Energy Input
16
ENERGY OUT
Products MASS
kg
kmol Cp
kJ/kmolK
HEAT
Q, kJ
Crude Trichloroethylene 4440 33.4 121.62 1.4×106
Calcium chloride 23% solution 8079 205 137.9 2.12×106
Heat loss
(vapor)
102 5.6 40626 (λ) 0.23×106
Total
energy
3.75×106
5.2. ENREGY BALANCE ACROSS THE CONDENSER
Vapor from the reactor which enters the condenser at 90°C with 1 bar pressure. At the more
mole fraction X=0.91 of condensate we get both liquid and vapor in T-x-y diagram
Liquid Temperature = 85°C,
We take cooling water enters at 30°C and leaving at 42°C
Table 5.1.2 Reactor Energy Output
Fig 5.2.1 Vertical Condenser
17
Product mixture Mol Percentage
%
Heat Capacity Cp
kJ/kmolK
Trichloroethylene 92.0 120.1
Dichlroethylene 2.63 112.1
Tetrachloroethylene 5.27 140
Heat capacity of mixture
Cp(mix)
121.62
Basis: 1 batch = 4 hours
Amount of Heat removed = Heat loss to reduction vapor temperature
from 90°c to 85°c
+
Heat loss due to condensation (latent heat removal)
Q1 = m Cp (mix) ΔT
= 33.4 x 121.62 x (90-85)
= 20311.7 KJ/batch
= 5077.9 kJ/hr
Q2 = m λ
= 4440 x 0.236
= 1047.84 kJ/batch
= 261.96 kJ/hr
Total Heat to be removed = Q1+ Q2 = 5339.86 kJ/hr
Table 5.2.1 Heat Capacity of Vapor
18
TUBE SIDE - COOLING MEDIA
Cold water available at 30°c and we assume it leaves at 42°c
Heat removed = mass flow rate of water x Heat capacity
Mass flow rate of cold water (m) = Q / (Cp ΔT)
= 5339.86x103 / [4.18 x (42-30)]
= 29.57 kg/hr
5.3. ENERGY BALANCE FOR COOLER
30°C
46°C
40°C
85°C
Fig 5.2.2 Temperature Profile
Fig 5.3.1 Cooler
19
The Crude Trichloroethylene condensate from condenser which enters a cooler at 85°C
(358K) and leaves at 40°C (313K).Here the cool water passed in counter current direction
is available at 30°C (303K) and leaves 46°C (319K)
Heat Duty
Q = m Cp ΔT
m – Molar flow rate of fluid (kmol/hr)
Cp- Specific Heat capacity of mixer (kJ/kmolK)
ΔT- Temperature differences (K)
Qh = mh Cph ΔT
= 8.349 x 121.627 x (358-313)
= 45695.89 kJ/hr
The required cool water mass flow rate
Qc = mc Cpc ΔT
We have to remove heat using cool water,
So, Qh = Qc
mc = Qh
Cpc ΔT
= 45695.89
4.18 (319-303)
= 683.25 kg/hr
20
5.4. ENERGY BALANCE IN FRACTIONATOR
In Fractionator
F2 feed rate = 942.88kg/hr
Composition (weight) (Mol)
TCE 93% 94.37%
PERC 7% 5.6%
Total = 7.074 kmol
Amount of heat in Feed
F (Hf) = [(6.676×120.1) + (0.398×147)] × (88)
= 74759.5 kJ/hr
88°C
87°C
110°
C
Fig 5.4.1 Distillation Column
21
Condenser section
V (Hv) = D (HD) + L (HL) + Qc
From the above equation,
Qc = V × λ
V = (1+R) ×D
D = 6.686 kmol/hr
V = (1+2.17) × 6.686 we know that (R = 2.17)
= 21.194 kmol/hr
V × λ = 21.194 × 40266
= 861027.44 kJ/hr
Qc = 86.103 × 104 kJ/hr
Cooling water requirement = Qc / [4.18 × (42-30)]
= 1.716 × 104
kg/hr
Reboiler section
Over all heat balance
F (Hf) + Qb = Qc + D (HD) + W (Hw)
Qb = Qc + D (HD) + W (Hw) – F (Hf)
Qb = 15305.79 + 86.103×104
– 74759.5 = 801576.29 kJ/hr
Steam available at 3 kg/cm2
(pressure steam temperature 135°C)
From steam table λV = 2172 kJ/kg
Steam requirement rate ms = Qb/ λV = 801576.29 / 2172
= 359 kg/hr
22
6. DESIGN OF MAJOR EQUIPEMENT
6.1. CONDENSER
PROCESS DESIGN
(I) Preliminary Calculations:
(a) Heat Balance
Vapor flow rate (G) = 8.349 kmoles/hr.
= 1110 kg/hr
= 0.3083 kg/s
Vapor Feed Inlet Temperature =900c.
Let Condensation occur under sub cooling conditions i.e. FT = 0.8
Condensate outlet temperature = 850C
Average Temperature = 87.50C
Latent heat of vaporization (λ) = {31.4(0.026) + 33.3(0.92) + 35.0(0.0527)}
= 33.4 kJ/mol
Qh = (molar flow rate of condensing vapor) x (latent heat of vapor)
Qh = heat transfer by the condensing vapor
Qh = 8.349 k-moles/hr x 33.4kJ/mol x (1000/3600)
= 275517.1 kJ/hr
= 77.945 kJ/s
10% overload is taken
Qh = 1.1 x 77.945 = 85.74 kJ QC = mass flow rate of cold × specific × t
fluid heat
23
QC = heat transfer by the cold fluid.
Assume: Qh = QC.
Inlet temperature of water = 30 0C.
Let the water be untreated water.
Outlet temperature of water (maximum) = 420C
t = 42-30= 120C
Cp = 4.187 kJ/kg K.
mc = 85.74 × 103
= 1 . 7 1 k g / s .
4.187x103x12
(b) LMTD Calculations
Assume, counter current
LMTD = (T1- t2) – (T2 - t1)
ln (T1- t2 )
(T2 - t1)
T1 = 87.5°C, T2 = 87.5°C, t1 = 30°C, t2 = 42°C; LMTD = 51.28°C
(ΔT) LMTD, corrected = 0.8 × 51.28°C = 41.02°C
Fig 6.1.1 Temperature Drop in Condenser
24
(C) Routing of fluids
organic vapors - Shell
side Liquid - Tube side
(D) Heat Transfer Area
(i) Qh = Qc =UA (ΔT) LMTD, corrected.
U= Overall heat transfer coefficient (W/m2
K)
Assume Ua =300W/m2K
A assumed = 85.74 ×103
300 × 41.02
= 6.886 m2
(ii) Select pipe size
Outer diameter of pipe (OD) = 1” = 0.0254 m
Inner diameter of pipe (ID) = 0.022m
Let length of tube = 2.5m
Let allowance = 0.05m
Heat transfer area of each tube (aheat – transfer) = π × OD × (Length – Allowance)
= π × 0.0254 × (2.5 – 0.05)
= 0.1954 m2
Number of tubes (Ntubes) = A assumed 6.886
=
a heat-transfer 0.1954
= 36 tubes
25
(iii) Choose Shell diameter:
Choose TEMA: P or S. 1” OD tubes in 1.25” □lar pitch.
1 - 1 Vertical Condenser
Ntubes (Corrected) = 48
Shell Diameter (Dc) =305 mm.
Acorrected =9.38 m2
Ucorrected = 220.25 W/m2K
(iv) Fluid velocity check:
(a) Vapor side – need not check
(b) Tube side
Flow area (atube) = apipe× Ntubes
Per pass
Ntube passes
a pipe = C.S of pipe = π(ID2)
4
atube = π (0.0254)2
× 48 = 0.0182 m2/pass
4 1
Velocity of fluid (Vpipe) vp = mpipe
in pipe ρpipe x atube
mpipe = mass –flow rate of fluid in pipe.
ρpipe = Density of fluid in pipe (water)
vp = 1.71 = 0.1 m/s
1000 × 0.0182
Fluid velocity check is satisfied
26
(II) Film Transfer Coefficient:
Properties are evaluated at tfilm :
tfilm = tv + tv + (t1+t2) 87.5 + 87.5 + (30+42)
2 = 2 = 74.62°C
2 2
a) Shell side
Reynolds‟s Number (Re) = 4 Г = 4 W µf µf (Ntubes)
2/3×L
= 4 0.3083 ×
0.0002 (48)2/3×2.5
= 185.5 m2
For vertical condenser
ho = 1.51 kf ρf g 0.33
4 W -0.33
µf2
L Nt0.667µf
ho = 1.51 (0.138)3 (1290)
2 (9.81)
0.33 185.5
-0.33
(0.2 × 10-3
)2
= 2510 W/m2
27
b) Tube side
Gt = mpipe
atube
Gt= Superficial mass velocity
Gt = 1.71 = 93.95 kg/m2s
0.0182
Re = (ID) Gt = 0.022 × 93.95 = 2583.6
μ 0.8 ×10 –3
Pr = μ Cp = 0.8 × 10 –3
× 4.187 × 10 3
= 5.79
K 0.578
hi (ID)
K = 0.023 (Re) 0.8 (Pr) 0.3
hi = inside – heat transfer coefficient
hi = 0.023 (2583.6) 0.8
(5.79) 0.3
× 0.578
0.022
hi = 549.20 W/m2K
Fouling factor
(Dirt –coefficient) = 0.003
=5.28 x 10-4 (W/m2K)-1
1 1 (OD) 1
= + + wall resistance + Fouling factor
U0 ho (ID) hi
Uo = overall heat –transfer coefficient
1 1 0.0254 1
= + × + 4.028x10-5 + 5.28x10-4
U0 1154.9 0.022 549.20
U0 = 325.85 W/m2K
U0 > UD
Therefore our assumption in overall heat transfer (UD) co-efficient is sufficient for design
28
(III) Pressure Drop Calculations
a) Tube Side
Re = 2583.6
f = 0.079 (Re)-¼
= 0.079 (2583.6) -¼
= 9.737 x 10 –3
f = friction factor
Pressure Drop along
the pipe length ( P) L = ( H)L × ρ × g
= 4fLVp2
× ρ × g
2g (ID)
= 4 × 9.737 × 10-3
×2.5 × 93.956 2
× 9.81
2 × 9.81 × 0.022 × 1000
= 1.96 kPa
Pressure Drop in the
end zones ( P)e = 2.5 ρ Vp2
= 2.5 x1000 × (93.95×10-2
) 2
=11.3 kPa 2 2
Total pressure drop
in pipe ( P) total = [1.96 + 11.3] 2 = 13.26 kPa < 70 kPa
b) Shell side, Kern‟s method
Baffle spacing (B) = Ds =305 mm
C′ = 3.1 x 10
–2 – 0.0254 = 0.0056m
PT = pitch = 3.1 x 10 –2
m
29
ashell = shell diameter × C′×B = 0.305 × 0.0056 × 0.305
PT 3.1× 10 –2
= 0.0168 m2
De = 4 PT×0.86 PT - 1 π (OD)2 = 4 (31 x 10
–3)
2 ×0.86 - π (0.0254)
2
2 2 4 2 8
(π do) π ( 0.0254)
2 2
= 16.0mm.
Gs= Superficial velocity in shell = mshell = 0.3083 = 18.351 kg/m
2s
ashell 0.0168
(NRe)s = Gs Dc = 18.351× 16 × 10 –3 = 35375
8.3 × 10-6
f = 1.87 (35375) –0.2
= 0.230
Shell side pressure
drop ( P)s = 4 f (Nb + 1) Ds Gs
2 g ] x 0.5
2 g De ρ v
Nb = 0
( P)s = 4 (0.230) (1) (0.305) (18.351)29.81 × 0.5
2 x 9.81 (16 x 10-3) × 4.57
= 6.461 KPa
(It very near to permissible pressure drop).
30
Shell Inner Diameter = 303mm
Shell thickness = 30mm
Inlet line Diameter = 80mm
Outlet Line Diameter = 18mm
Head = Elliptical type
Tube Inner Diameter = 25.4mm
Tube Length = 2500mm
Fig 6.1.2 Model of Vertical Condenser
31
6.2. DISTILLATION COLUMN
87°C D = 6.686 kmoles xD = 0.99
Enrich.
section
F= 7.074k.moles xF = 0.931 TF =90°C Stripping
Section
100°C
Basis: 1 hour operation
Glossary of notations used
Total
Reboiler
W = 0.38kmol Xw= 0.03
F = molar flow rate of feed, kmol/hr D = molar flow rate of distillate, kmol/hr
W = molar flow rate of residue, kmol/hr.
xF = mole fraction of TCE in liquid
xD = mole fraction of TCE in distillate xW = mole fraction of TCE in residue
Fig 6.2.1 Distillation Column
Design
32
Rm = minimum reflux ratio R = actual reflux ratio
L= molar flow rate of liquid in the enriching section, kmol/hr
G = molar flow rate of vapor in the enriching section, kmol/hr
L = molar flow rate of liquid in stripping section, kmol/hr
G = molar flow rate of vapor in stripping section, kmol/hr
M‟ = average molecular weight of feed, kg/kmol
q = Thermal condition of feed
Feed = Saturated liquid at its bubble point, Temperature= 90°C
M‟ = 132.95 kg/kmol
Relative volatility which predicts the order to separate binary mixture using distillation column
process,
Thus the relative volatility (α) of this mixture is obtained from data page (Perry‟s Hand book)
α = 2.67@ 760mmHg, α = y (1-x)/x (1-y)
Vapor liquid equilibrium data
The operating line for the rectification section is constructed as follows. First the desired top
product composition is located on the VLE diagram, and a vertical line produced until it
X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Y 0.228 0.40 0.533 0.64 0.727 0.80 0.86 0.914 0.96 1
Fig 6.2.2 T-x-y Diagram
33
intersects the diagonal line that splits the VLE plot in half. A line with slope R/(R+1) is then
drawn from this intersection point as shown in the diagram below.
R is the ratio of reflux flow (L) to distillate flow (D) and is called the reflux ratio and is a
measure of how much of the material going up the top of the column is returned back to the
column as reflux.
xD Rm+1 = 0.66
Rm+1 = xD 0.994
= = 1.5 0.66 0.66
Rm = 1.5 – 1.00 = 0.5
R= 1.5 Rm = 0.75
xD 0.994
= = 0.568
R+1 0.75 +1
Number of trays from graph =12
L = RD = 0.568x6.686 = 3.797 k-moles
Fig 6.2.3 V-L-E Diagram
34
G =L+D = 1.838 + 6.686
= 10.483 k-moles q=1 (Feed is saturated liquid)
L = L+qf = 3.797 + 1(7.074)
= 10.871 k-moles
G = G+ (q –1) F = 8.475 +0
= 10.483 k-moles
Fig 6.2.4 McCabe Thiele Diagram
35
PROPERTIES
Enriching section Stripping section
Top Bottom Top Bottom
Liquid (k.moles/hr)
Liquid (kg/hr)
Vapor
(k.moles/hr)
Vapor (kg/hr)
x
y
T liquid (oC)
T vapor(oC)
ρvapor(kg/m3)
ρliquid (kg/m3)
(L/G)(ρg/ρL)0.5
liq (dyn/cm)
µ vapor
µ liq
Dvapor (m2/s)
Dliquid (m2/s)
3.797
499.30
10.483
1378.52
0.994
0.99
87
87.6
4.57
1460
0.0202
28.7
00038
0.27
0.0599
1.688x10-9
3.797
504.81
10.483
1393.71
0.952
0.95
89
89.7
4.51
1480
0.0199
28.7
0.0038
0.271
0.0059
1.688x10-9
10.871
1467.58
10.483
1415.2
0.93
0.932
89.5
90.3
4.46
1515
0.056
26.3
0.0038
0.449
0.148
1.952x10-9
10.871
1804.58
10.483
1740.18
0.036
0.036
100
101.5
5.1
1546
0.0595
26.3
0.0038
0.4491
0.149
1.952x10-9
Table 6.2.1 Properties of Vapor and Liquid in Distillation
column
36
AVERAGE CONDITIONS AND PROPERTIES
Properties Enriching section Stripping section
Liquid (k-moles/hr) (kg/hr)
3.797 502.05
10.871 1636.08
Vapor ( k-moles/hr)
(kg/hr)
10.483
1386.115
10.483
1577.69
Tliq (°c )
88
94.75
Tvapor (°c )
88.65 95.9
ρliq (kg/m3)
1470 1530.5
ρvapor (kg/m3)
4.54 2.483
ENRICHING SECTION
VOLUMETRIC FLOW RATE OF VAPOUR
Assume ideal gas behavior
V = nRT/P
= 3.787 × 0.08206 x 361.5/1 × 3600
V=0.313 m3/s
Assuming vapor velocity = 0.5 m/s
Net cross sectional area of top column =Vol .flow rate / vap. Velocity
= 0.313/0.5
An = 0.624 m2
Table 6.2.2 Average Conditions and Properties
37
Area of the column, Ac=An/0.88
=0.624/0.88
=0.81m2
Ac= (/4) x Dc2
0.81=0.785Dc2
Dc = 1.015 m
STRIPPING SECTION
VOLUMETRIC FLOW RATE OF VAPOUR:
Assume ideal gas behavior
V = nRT/P
= 10.871 x 0.08206 × 367.75/1 × 3600
V= 0.278 m3/s
Vapor velocity =0.5 m/s
Net cross sectional area of top column =Vol .flow rate / vap. Velocity
=0.278/0.5
An = 0.556 m2
Area of the column, Ac=An/0.88
=0.556/0.88
=0.632 m2
Ac= (/4) xDc2
0.632=0.785Dc2
Dc = 0.897 m
38
CROSS SECTIONAL AREA OF DOWN COMER
Ad = Ac - An
=0.81 – 0.556
Ad =0.254 m2
TOWER HEIGHT:
= ((Actual trays – tray for reboiler + tray for condenser) +2) x Tray spacing
Actual trays = (NTP/Tray overall efficiency)
= (12/ 0.17 – 0.616 log (μavg))
= (12 / 0.474) since, μavg = 0.32 cP
= 25.27 = 26
= ((26-1+0) +2) × Tray spacing
= 27 × Tray spacing
=27 × 0.5
Column height =13.5 m
FEED ENTERING ZONE
Feed enter in between the enriching to stripping section. From the above parameters feed
should enter in the 5th
tray which find out from the actual tray calculation and help of graph
chart.
39
TRAY HYDRAULICS (TOP & BOTTOM)
Sieve tray are used
1) Plate spacing = 0.5 m
2) Hole diameter = 0.006 m
3) Hole pitch = 3 x Hole diameter =0.0018 m
4) Tray thickness =0.6 × 0.006=0.0036 m
Ratio of hole area to perforated area = (Ah / Ap)
= 0.9 × (6/18) 2
= 0.1
WEIR HEIGHT
For normal pressure weir height lies between
40-50 mm
Let weir height = 0.05 m
WEIR LENGTH
Weir length = 0.75 × Dc
= 0.75 × 0.879
Weir length = 0.66 m
40
7. COST ESTIMATION AND ECONOMICS
Cost of TCE plant of capacity 5430 TPY in 1968 Rs. 3.26×108
Therefore cost of 7602 TPY in 2011 is: C1 = C2 (Q1/Q2)0.6
= 3.26×108 (7602/5430)
0.6
= Rs.3.99 × 108 Chemical Engineering Plant Cost Index
Cost index in 1968 = 112
Cost index in 2011 = 511 (from chemical magazines) Thus, Present cost of Plant = (original cost) × (present cost index)/(past cost index)
= (Rs.3.99 × 108) × (511/112) = Rs. 18.20×108
i.e., Fixed Capital Cost (FCI) = Rs. 18.20×108 Estimation of Capital Investment Cost I. Direct Costs: material and labor involved in actual installation of complete facility (70-
85% of fixed-capital investment)
a) Equipment + installation + instrumentation + piping + electrical + insulation
+ painting (50-60% of Fixed-capital investment) 1. Purchased equipment cost (PEC): (15-40% of Fixed-capital investment)
Consider purchased equipment cost = 25% of Fixed-capital investment
i.e., PEC = 25% of 18.20×108
= 0.25 × 18.20×108
= Rs. 4.551×108
2. Installation, including insulation and painting: (25-55% of purchased equipment
cost.)
Consider the Installation cost = 40% of Purchased equipment cost
= 40% of 4.551×108
= 0.40 ×4.551×108
= Rs.1.820×108
41
3. Instrumentation and controls, installed: (6-30% of Purchased equipment cost.) Consider the installation cost = 20% of Purchased equipment cost
= 20% of ×4.551x108
= 0.20 ×4.551×108
= Rs. 0.9102×108 4. Piping installed: (10-80% of Purchased equipment cost)
Consider the piping cost = 40% Purchased equipment cost
= 40% of Purchased equipment cost
= 0.40 ×4.551×108
= Rs. 1.8204×108
5. Electrical, installed: (10-40% of Purchased equipment cost)
Consider Electrical cost = 25% of Purchased equipment cost
= 25% of 4.551×108
= 0.25 ×4.551×108
= Rs. 1.1377×108
B. Buildings, process and Auxiliary: (10-70% of Purchased equipment
cost)
Consider Buildings, process and auxiliary cost = 40% of PEC
= 40% of 4.551×108
= 0.40 ×4.551×108
= Rs. 1.8104×108
C. Service facilities and yard improvements: (40-100% of Purchased equipment
42
cost)
Consider the cost of service facilities and yard improvement = 60% of PEC
= 60% of 4.551×108
= Rs. 2.7306×108
D. Land: (1-2% of fixed capital investment or 4-8% of Purchased equipment cost)
Consider the cost of land = 6% PEC
= 6% of 4.551×108
= 0.06 ×4.551×108
= Rs. 0.2730×108
Thus, Direct cost = Rs. 15.0586×108 --------- (82.74% of FCI)
II. Indirect costs: expenses which are not directly involved with material and labor of actual
installation of complete facility (15-30% of Fixed-capital investment)
A. Engineering and Supervision: (5-30% of direct costs) Consider the cost of engineering and supervision = 10% of Direct costs
i.e., cost of engineering and supervision = 10% of 15.0519 ×108
= Rs. 1.5058×108
B. Construction Expense and Contractor‟s fee: (6-30% of direct costs)
Consider the construction expense and contractor‟s fee = 10% of Direct costs
i.e., construction expense and contractor‟s fee = 10% of 15.0586×108
= 1.5059×108
C. Contingency: (5-15% of Fixed-capital investment) Consider the contingency cost = 10% of Fixed-capital investment
i.e., Contingency cost = 10% of 18.20×108
= 0.12× 18.20×108
= Rs. 1.820×108
Thus, Indirect Costs = Rs. 5.1942×108 --------- (28.54% of FCI)
43
III. Fixed Capital Investment: Fixed capital investment = Direct costs + Indirect costs
= (15.0586×108) + (5.1942×108)
i.e., Fixed capital investment = Rs. 20.253×108 IV. Working Capital: (10-20% of Fixed-capital investment)
Consider the Working Capital = 15% of Fixed-capital investment
i.e., Working capital = 15% of 20.253×108
= 0.15 × 20.253×108
= Rs. 3.037×108
V. Total Capital Investment (TCI): Total capital investment = Fixed capital investment + Working capital
= (20.253×108) + (3.037×108)
i.e., Total capital investment = Rs. 23.291×108
Estimation of Total Product cost I. Manufacturing Cost = Direct production cost + Fixed charges + Plant overhead
cost.
A. Fixed Charges: (10-20% total product cost) i. Depreciation: (depends on life period, salvage value and method of calculation-
about 13% of FCI for machinery and equipment and 2-3% for Building Value for Buildings)
Consider depreciation = 13% of FCI for machinery and equipment and 3% for Building
Value for Buildings)
i.e., Depreciation = (0.13×20.253×108) + (0.03×1.8104×108) = Rs. 2.687×108
ii. Local Taxes: (1-4% of fixed capital investment)
Consider the local taxes = 3% of fixed capital investment
i.e. Local Taxes = 0.03×20.253×108
44
= Rs. 0.607×108
iii. Insurances: (0.4-1% of fixed capital investment)
Consider the Insurance = 0.7% of fixed capital investment
i.e. Insurance = 0.007×20.253×108
= Rs. 0.1417×108
iv. Rent: (8-12% of value of rented land and buildings)
Consider rent = 10% of value of rented land and buildings
= 10% of ((0.2730×108) + (1.8104×108))
Rent = Rs. 0.2083x108
Thus, Fixed Charges = Rs. 3.644×108
B. Direct Production Cost: (about 60% of total product cost) Now we have Fixed charges = 10-20% of total product charges – (given)
Consider the Fixed charges = 15% of total product cost
Total product charge = fixed charges/15%
Total product charge = 3.644×108/15%
Total product charge = 3.644×108/0.15
Total product charge (TPC) = Rs. 24.29×108 i. Raw Materials: (10-50% of total product cost) Consider the cost of raw materials = 25% of total product cost Raw material cost = 25% of TPC
= 0.25×24.29×108
Raw material cost = Rs. 6.0725×108 ii. Operating Labor (OL): (10-20% of total product cost)
Consider the cost of operating labor = 12% of total product cost
Operating labor cost = 12% of 24.29×108 = 0.12×24.29×10
8
Operating labor cost = Rs. 2.914×108 iii. Direct Supervisory and Clerical Labor (DS & CL): (10-25% of OL) Consider the cost
45
for Direct supervisory and clerical labor = 12% of OL
Direct supervisory and clerical labor cost = 12% of 2.914×108
= 0.12×2.914×108
Direct supervisory and clerical labor cost = Rs. 0.3497×108
iv. Utilities: (10-20% of total product cost)
Consider the cost of Utilities = 12% of total product cost
Utilities cost = 12% of 24.29×108
= 0.12×20.253×108
Utilities cost = Rs. 2.43×108 v. Maintenance and repairs (M & R): (2-10% of fixed capital investment)
Consider the maintenance and repair cost = 5% of fixed capital investment
i.e. Maintenance and repair cost = 0.05×20.253×108
= Rs.1.0126×108
vi. Operating Supplies: (10-20% of M & R or 0.5-1% of FCI)
Consider the cost of Operating supplies = 15% of M & R Operating supplies cost
= 15% of 1.085×108
= 0.15×1.0126×108
Operating supplies cost = Rs. 0.1518×108 vii. Laboratory Charges: (10-20% of OL)
Consider the Laboratory charges = 15% of OL
Laboratory charges = 15% of 2.914×108 = 0.15×2.914×108
Laboratory charges = Rs. 0.6204×108 viii. Patent and Royalties: (0-6% of total product cost) Consider the cost of Patent and royalties = 4% of total product cost
Patent and Royalties = 4% of 24.29×108
= 0.03×24.29×108
Patent and Royalties cost = Rs. 0.7287×108
46
Thus, Direct Production Cost = Rs. 14.279×108 -------- (59% of TPC)
C. Plant overhead Costs (50-70% of Operating labor, supervision, and
maintenance or 5-15% of total product cost); includes for the following: general plant
upkeep and over head , payroll overhead, packaging, medical services, safety and
protection, restaurants, recreation, salvage, laboratories, and storage facilities.
Consider the plant overhead cost = 60% of OL, DS & CL, and M & R
Plant overhead cost = 60% of [(2.914×108) + (0.3497×108) + (1.0126×108)]
Plant overhead cost = Rs. 2.565×108
Thus, Manufacture cost = Direct production cost + Fixed charges + Plant overhead costs.
Manufacture cost = (24.29×108) + (3.644×108) + (2.565×108) Manufacture cost = Rs.30.499×10
8
II. General Expenses = [Administrative costs + distribution and selling costs + Research and development costs] A. Administrative costs: (2-6% of total product cost) Consider the Administrative costs = 5% of total product cost
Administrative costs = 0.05 × 24.29×108
Administrative costs = Rs. 1.2145×108 B. Distribution and Selling costs: (2-20% of total product cost); includes costs for sales
offices, salesmen, shipping, and advertising.
Consider the Distribution and selling costs = 15% of total product cost
Distribution and selling costs = 15% of 24.29×108
Distribution and selling costs = 0.15 × 24.29×108
Distribution and Selling costs = Rs. 3.464×108
C. Research and Development costs: (about 5% of total product cost)
Consider the Research and development costs = 5% of total product cost
Research and Development costs = 5% of 24.29×108
Research and development costs = 0.05 × 24.29×108
47
Research and Development costs = Rs. 1.2145×108 D. Financing (interest): (0-10% of total capital investment)
Consider interest = 5% of total capital investment
i.e. interest = 5% of 23.291×108 = 0.05×23.291×108
Interest = Rs. 1.1645×108
Thus, General Expenses = Rs. 7.0575×108
IV. Total Product cost = Manufacture cost + General Expenses
= (30.499×108) + (7.0575×108)
Total product cost = Rs. 37.5565×108
V. Gross Earnings/Income: Wholesale Selling Price of TCE per kg = Rs.60 Total Income = Selling price × Quantity of product manufactured
= 60 x 0.7602 x 108
Total Income = Rs. 45.612×108 Gross income = Total Income – Total Product Cost
= (45.612×108) – (37.5565×108)
Gross Income = Rs. 8.056×108
Let the Tax rate be 45% (common)
Net Profit = Gross income - Taxes
= Gross income× (1- Tax rate)
Net profit = 8.056×108(1- 0.45)
= Rs. 4.431×108
Rate of Return Rate of return = Net profit×100/Total Capital Investment
Rate of Return = 4.431×108×100/ (23.291×108)
Rate of Return = 19.02%
48
Break-even Analysis Data available
Annual Direct Production Cost = Rs. 24.29×108
Annual Fixed charges, overhead and general expenses = Rs. 3.644×108
Total Annual sales = Rs. 148.24×108 Wholesale Selling Price MEK per ton. = Rs. 60000
Direct production cost per ton of MEK = (3.644×108)/ (148.24×108/60000) = Rs. 1474.9 per ton Let „n‟ TPA be the break even production rate. Number of tons needed for a break-even point is given by
(3.644×108) + (1474.9 ×n) = (60000×n) => n = 6226.39 tons/year n = 1.7199 tons/day
Hence, the break-even production rate is 1.7199 TPD or 8 .2% of the considered plant
capacity.
49
8. PLANT LAYOUT
The location of the plant can have a crucial effect on the overall profitability of a project,
and the scope for future expansion. Many factors must be considered when selecting a suitable
plant site. The principal factors are:
• Location, with respect to the marketing area
• Raw material supply
• Transport facilities
• Availability of labor
• Availability of suitable land
• Environmental impact and effluent disposal
• Local community consideration
• Climate
• Political and strategic consideration
Raw material availability
The source of raw material is one of the most important factors influencing the selection of a
plant site. This is particularly true if large volumes of raw material are consumed, because
location near the raw material source permits considerable reduction in transportation and
storage charge. Attention should be given to the purchased price of the raw material , distance
from the source of supply , fright or transportation expenses availability and reliability of supply
purity of the material and storage requirements.
Markets
The location of markets or intermediate centers affects the cost of product distribution and
the time required for shipping. Proximity to the major markets is an important consideration in
the selection of plant site , because the buyer usually finds it advantages to purchase from nearby
sources. It should be noted that markets are needed for by products as well as major final
products.
50
Energy availability
Power and steam requirements are high in most industrial plants , and fuel is ordinary
required to supply these utilities. Consequently power and fuel can be one major factor in the
choice of a plant site. Electrolytic processes require a cheap source of electricity and plants using
electrolytic processes are often located near hydroelectric installations. If the plant requires large
quantities of coal or oil, location near a source of fuel supply may be essential for economic
operation. The local cost of power can help determine whether power should be purchased or
self-generated.
Climate
If the plants is located in a cold climate, costs may be increased by the necessity for
construction of protective shelters around the process equipment, and special cooling towers or
air –conditioning equipment may be required if the prevailing temperature are high. Excessive
humidity or extremes of hot or cold weather can have a serious effect on the economic operation
of plant and these factors should be examined when selecting a plant site.
Transportation facilities
Water, railroads and high and ways are the common means of transportation used by major
industrial concerns. The kind and amount of products and raw materials determine the most
suitable type of transportation facilities. In any case, careful attention should be given to local
freight rates and existing railroad lines. The proximity to railroad centers and possibility of canal,
river, lake or ocean transport must be considered. Motor trucking facilities are widely used and
serve as a useful supplement to rail and water facilities. If possible ,the plant site should have
access to all three types of transportation, and certainly, at least two types should be available .
There is usually need for convenient air and rail transportation facilities between the plant and
the main company headquarters, and effective transportation facilities for the plant personnel are
necessary.
Water supply
The process industries use large quantities of water for cooling, washing steam generation and
as raw material. The plant therefore must be located where a dependable supply of water is
51
available. A large river must be located where a dependable supply of water may be satisfactory
if the amount of water required is not too great. The level of the existing water table can be
checked by constancy of the water table and the year-round capacity of local rivers or lakes
should be obtain. It the water supply shows seasonal fluctuations, it may be to construct a
reservoir or to several standby wells. The temperature, mineral content, silt or sand content,
bacteriological content, and cost for supply and purification treatment must also be considered
when choosing a water supply.
Waste disposal
In recent years, many legal restrictions have been placed on the methods for disposing of waste
material from the process industries. The site selected for disposing of a plant should have
adequate capacity and facilities for correct waste disposal. Even though a given area has
minimal restrictions on pollution. It should not be assumed that this condition will continue to
exist. In choosing a plant sit, the permissible tolerance levels for various methods of waste
disposal should be considered carefully and attention should be given to potential for additional
waste-treatment facilities.
Labor supply
The type and supply of labor available in the vicinity of a proposed plant site must be
examined. Consideration should be given to prevailing pay scales, restrictions on number of
hours worked per week, competing industries that can cause dissatisfaction or high turnover rates
among the workers, and variations in the skill and productivity of the workers.
Taxation and legal restrictions
State and local tax rates on property income, unemployment insurance and similar items
vary from form one location to another. Similarly, local regulations on zoning, building codes,
nuisance aspects and transportation can have a major influence on the final choice of a plant site.
In fact, zoning difficulties and obtaining the many required permits can often be much more
important in terms of cost and time delays than many of factors discussed in the preceding
sections.
52
Site characteristics
The characteristics of the land at a proposed plant site should be examined carefully. The
topography of the tract of land and the soil structure must be considered, since either of both may
have a pronounced effect on construction cost or living conditions. Future changes may make it
desirable or necessary to expand the plant facilities. Therefore , even through no immediate
expansion is planned, a new plant should be constructed at a location where additional space is
available.
Flood and fire protection
Many industrial plants are located along rivers near large bodies of water, and risks of flood
of flood or hurricane damage. Before selecting a plant site, the regional history of natural events
of this type should be examined and the consequences of such occurrence considered. Protection
from losses by fire is another important factor in selecting a plant location. In case of major fire,
assistance from outside fire department should be available. Fire hazards in the immediate area
surrounding the plant site must not be overlooked.
Community factors
The character and facilities of a community can have quite an effect on the location of
plant. If a certain minimum number of facilities for satisfactory living of plant personnel do not
exist, it often becomes a burden for the plant to subsidize such facilities. Cultural facilities of the
community are important to sound growth. Churches libraries, schools, civic theaters, concert
associations and other similar groups, if active and dynamic, do much to make a community
progressive. The problem of recreation deserves special consideration. The efficiency, character,
and history of both state and local government should be evaluated. The existence of low taxes is
not in itself a favorable situation unless the community is already well developed and relatively
free of debt.
53
PLANT LAYOUT
The economic construction and operation of a process unit will depend on how well the
plant equipment specified on the process flow sheet and laid out.
The principal factors to be considered are:
1. Economic consideration: construction and operation cost.
2. The process requirement
3. Convenience of operation
4. Convenience of maintenance
5. Safety
6. Future expansion
7. Modular construction
COSTS
The cost of construction can be minimized by adopting a layout that gives shortest run of
connecting pipes between equipment, and adopting the least amount of structural steel work.
However, this will not necessarily be the best arrangement for operation and maintenance.
PROCESS REQUIREMENT
All the required equipments have to be placed properly within process. Even the
installation of the auxiliaries should be done in such a way that it will occupy the least space.
OPERATION
Equipment that needs to have frequent operation should be located convenient to the
control room. Valves, sample points, and instruments should be located at convenient position
and height. Sufficient working space and headroom must be provided to allow easy access to
equipment.
54
MAINTENANCE
Heat exchangers need to be sited so that the tube bundles can be easily withdrawn for
cleaning and tube replacement. Vessels that require frequent replacement of catalyst or packing
should be located on the outside of buildings. Equipment that requires dismantling for
maintenance, such as compressors and large pumps, should be placed under cover.
SAFETY
Blast walls may be needed to isolate potentially hazardous equipment, and confine the
effects of an explosion. At least two escape routes for operator must be provided from each level
in the process building.
PLANT EXPANSION
Equipment should be located so that it can be conveniently tied in with any future
expansion of the process. Space should be left on pipe alleys for future needs, service pipes
oversized to allow for future requirements.
MODULAR CONSTRUCTION
In recent years, there has been a move to assemble sections of the plant at the
manufacturer site. These modules will include the equipment, structural steel, piping and
instrumentation. The modules then transported to the plant site, by road or sea.
55
H e a l t h 2
F i r e 1
R e a c t i v i t y 0
P e r s o n a l H
P r o t e c t i o n
9. Material Safety Data Sheet
Trichloroethylene MSDS
2 0
Product Name: Trichloroethylene
Catalog Codes: SLT3310, SLT2590
CAS#: 79-01-6
RTECS: KX4560000
TSCA: TSCA 8(b) inventory: Trichloroethylene
CI#: Not available.
Synonym:
Chemical Formula: C2HCl3
Composition and Information on Ingredients
Composition
CAS #
% by Weight
Name
Trichloroethylene
79-01-6
100
Hazards Identification
Potential Acute Health Effects: Hazardous in case of skin contact (irritant, permeator), of
eye contact (irritant), of ingestion, of inhalation.
Potential Chronic Health Effects:
CARCINOGENIC EFFECTS: Classified + (PROVEN) by OSHA. Classified A5 (Not
suspected for human.) by ACGIH. MUTAGENIC EFFECTS: Not available. TERATOGENIC
EFFECTS: Not available. DEVELOPMENTAL TOXICITY: Not available. The substance is
toxic to kidneys, the nervous system, liver, heart, upper respiratory tract. Repeated or
prolonged exposure to the substance can produce target organs damage.
Fig 9.1 Hazard Identity
Symbol
56
First Aid Measures
Eye Contact
Check for and remove any contact lenses. Immediately flush eyes with running water for at
least 15 minutes, keeping eyelids open. Cold water may be used. Do not use an eye
ointment. Seek medical attention.
Skin Contact
After contact with skin, wash immediately with plenty of water. Gently and thoroughly wash
the contaminated skin with running water and non-abrasive soap. Be particularly careful to
clean folds, crevices, creases and groin. Cover the irritated skin with an emollient. If irritation
persists, seek medical attention. Wash contaminated clothing before reusing.
Serious Skin Contact
Wash with a disinfectant soap and cover the contaminated skin with an anti-bacterial cream.
Seek medical attention.
Inhalation
Allow the victim to rest in a well ventilated area. Seek immediate medical attention.
Serious Inhalation
Evacuate the victim to a safe area as soon as possible. Loosen tight clothing such as a
collar, tie, belt or waistband. If breathing is difficult, administer oxygen. If the victim is
not breathing, perform mouth-to-mouth resuscitation. Seek medical attention.
Ingestion
Do not induce vomiting. Loosen tight clothing such as a collar, tie, belt or waistband. If
the victim is not breathing, perform mouth-to-mouth resuscitation. Seek immediate
medical attention.
Serious Ingestion Not available.
57
Fire and Explosion Data
Flammability of the Product: May be combustible at high temperature.
Auto-Ignition Temperature: 420°C (788°F)
Flash Points: Not available.
Flammable Limits: LOWER: 8% UPPER: 10.5%
Products of Combustion: These products are carbon oxides (CO, CO2), halogenated
compounds.
Fire Hazards in Presence of Various Substances: Not available.
Explosion Hazards in Presence of Various Substances:
Risks of explosion of the product in presence of mechanical impact: Not available.
Risks of explosion of the product in presence of static discharge: Not available.
Fire Fighting Media and Instructions:
SMALL FIRE: Use DRY chemical powder.
LARGE FIRE: Use water spray, fog or foam. Do not use water jet.
Special Remarks on Fire Hazards: Not available.
Special Remarks on Explosion Hazards: Not available
Accidental Release Measures
Small Spill: Absorb with an inert material and put the spilled material in an appropriate waste
disposal.
Large Spill: Absorb with an inert material and put the spilled material in an appropriate waste
disposal. Be careful that the product is not present at a concentration level above TLV. Check
TLV on the MSDS and with local authorities.
58
Handling and Storage
Precautions
Keep locked up Keep away from heat. Keep away from sources of ignition. Empty
containers pose a fire risk; evaporate the residue under a fume hood. Ground all
equipment containing material. Do not ingest. Do not breathe gas/fumes/ vapor/spray.
Wear suitable protective clothing In case of insufficient ventilation, wear suitable
respiratory equipment if ingested, seek medical advice immediately and show the
container or the label. Avoid contact with skin and eyes
Storage
Keep container dry. Keep in a cool place. Ground all equipment containing material.
Carcinogenic, teratogenic or mutagenic materials should be stored in a separate locked safety
storage cabinet or room.
Exposure Controls/Personal Protection
Engineering Controls:
Provide exhaust ventilation or other engineering controls to keep the airborne concentrations of
vapors below their respective threshold limit value. Ensure that eyewash stations and safety
showers are proximal to the work-station location.
Personal Protection:
Splash goggles. Lab coat. Vapor respirator. Be sure to use an approved/certified respirator or
equivalent. Gloves.
Personal Protection in Case of a Large Spill:
Splash goggles, Full suit, Vapor respirator, Boots, Gloves. A self contained breathing
apparatus should be used to avoid inhalation of the product. Suggested protective
clothing might not be sufficient; consult a specialist BEFORE handling this product.
Exposure Limits: TWA: 50 STEL: 200 (ppm) from ACGIH (TLV) TWA: 269 STEL: 1070
(mg/m3) from ACGIH Consult local authorities for acceptable exposure limits.
59
Stability and Reactivity Data
Stability: The product is stable.
Instability Temperature: Not available.
Conditions of Instability: Not available.
Incompatibility with various substances: Not available.
Corrosivity: Extremely corrosive in presence of aluminum. Non-corrosive in presence of glass.
Special Remarks on Reactivity: Not available.
Special Remarks on Corrosivity: Not available.
Polymerization: No
Transport Information
DOT Classification: CLASS 6.1: Poisonous material.
Identification: Trichloroethylene : UN1710 PG: III
Special Provisions for Transport: Not available
60
10. CONCLUSION
In Trichloroethylene manufacturing process, the above explained dehydrochlorination with
Calcium Hydroxide is the energy efficient technique that going in DCW limited. Because the
pyrolysis of Tetrachloroethane requires huge energy. They are only one Industry producing TCE
with 7600 TPY the way of lime process. Also according to the demand of TCE, they are
expanding their production to 10860 TPY. Because, the modern chemical derived from TCE for
the Refrigerant production instead of CFC. An industry, they are performing the condensation of
crude Trichloroethylene vapor using open type co-current vertical condenser with shell side
condensation that vapor product resulting from dehydrochlorination of Tetrachloroethane. I
suggested that present condenser can be replaced by closed type counter current vertical
condenser. This kind of modification will minimize the cold water requirement to 60% of current
status. This will reduce the cooling tower load. Thus above modification will feasible and
process becomes more effective.
61
11. BIBLIOGRAPHY
11.1. BOOKS CONSULTED
(1) M. Gopala Rao and Marshall Sittig, “Dryden‟s Outlines of Chemical Technology”,
3rd Ed., East-West press.[1990]
(2) Kirk-Othmer, “Encyclopaedia of Chemical Technology”, 5th Ed, Volume-1, John
Wiley & Sons Inc..[1971]
(3) I.Mukhlyonov & I.Furmer, “The most important industrial chemical process” part-2
MIR publishers.[1987]
(4) R. H. Perry and Don W. Green, “Perry‟s Chemical Engineers‟ Hand Book”, 6th and
7th Ed. Mc-Graw Hill International edition,[1989, 1993]
(5) OLEF A. Hougen, Kenneth M. Watson, Roland A. Ragatz “Chemical Process Principles”,
2nd
edition., CBS Publishers & Distributors, New Delhi.[1986]
(6) R. K. Sinnott, “Coulson and Richardson‟s Chemical Engineering Series, volume-6,
Chemical Equipment Design” 3rd Ed., Butter Worth-Heinemann, Page No: 828-855.
(7) Joshi M. V., “Process Equipment Design”, 2nd Ed., Mc-Millan India Ltd,[2000]
(8) Max S. Peters and Klaus Timmerhaus, “Process Plant Design and Economics For Chemical
Engineers”, 3rd Ed., Mc-Graw Hill Book Company, Page No: 207-208, 484-485[1996].
(9) B.C Bhattacharya, “Chemical equipment Design”, Chemical Engineering Education
Development Centre.[1998]
(10) L.E. Brownell and E.H. Young, “Process Equipment Design”, John Wiley & SonsInc. New
York,[2001]
62
11.2. WEB LINKS
www.wikipedia.com/1,1,2trichloroethyene+(data page)&Wikipedia%20the+free
encyclopedia.htm
http://www.cheresources.com/invision/topic/941-vertical-condenser-design/
http://www.cambridgesoft.com/databases/details/?fid=140
www.chemcad.com
http://www.freepatentsonline.com/3949009.html
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