snk final project
TRANSCRIPT
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CHAPTER 1
INTRODUCTION
1.1 HISTROY
In the year 1882, Pentaerythritol was discovered by product of the reaction
hydroxide and impure formaldehyde. The name Pentaerythritol which was assigned
to this compound was derived from erithritol to indicate the presence of 4 - hydroxyl
groups and the prefix Penta to show that there are 5 carbon atoms in the
molecules.
Pentaerythritol is a white crystalline compound. The high melting point, slight
solubility in water and the ready reactivity of its 4 hydroxide groups have been
attributed to the compact symmetric structure of the molecules. It is an optically
inactive compound, resembling cane sugar in appearance and has a sweet taste
characteristic of polyols.
Pentaerithritol is nonhygroscopic and stable in air and sublimes slowly on
heating. It is moderately soluble in cold water, quite soluble in hot water, and has
only limited solubility in organic liquids. It is used in paints, varnishes industry and
the production of resins. Pentaerithritol is increasingly used in resins manufacturing
mainly because of its desirable characteristics and price stability.
1.2 APPLICATION OF PENTAERITHRITOL
It is used to manufacture of Paints & Varnishes.
It is used to manufacture of Pentaerithritol Nitrate (P.E.T.N).
It is used to manufacture of Plasticizers & Lubricants.
Pentaerithritol combination with metal salts used as heat stabilizers.
Acrolin prepared from Pentarithritol used for electrical insulation,surface
coating films and fibers
.
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CHAPTER 2
AIM AND SCOPE
The objective of the project work is to study the manufacture of Pentaerithritol for
600 tons per annum.
To solve the material balance and energy balance for the process
To design the reactor and vacuum tray dryer involved in the process
To study the plant layout, safety, health and environment aspects
To estimate the cost and payback period of the process
The scope of the project is to select the method for the manufacturing of
pentaerithritol for the following features mentioned below
Less energy consumption.
High purity.
High yield / conversion.
Recovery of by product as sodiumformate.
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CHAPTER 3
PROCESS DESCRIPTION AND DESIGN
3.1. METHODS OF PRODUCTION
3.1.1 Using soda ash as condensing agent
Raw Materials Required
a) Formaldehyde
b) Ca (OH) 2 solution
c) Acetaldehyde
d) Soda ash
3.1.2 Using Sodium hydroxide as condensing agent
Raw materials required
a) Formaldehyde
b) Formic acid
c) Acetaldehyde
d) Sodium hydroxide
3.1.3 Slection of method:
Using NaOH as condensing agent
4 HCHO + CH3CHO + NaoH (CH2OH) 4 C+ HCOONa
Pentaerithritol
Justification of Process Selection
Yield 85-90%
Less energy consumption
Easy removal Recovery of by product as sodiumformate
High yield / conversion
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3.1.4 Process description
A solution of formadehyde (30%) is added to NaOH solution. The temperature is
maintained at 30oC by suitable agitation. liquid acetaldehyde (99%) is added slowly
added under the surface of formaldehyde alkali solution. External cooling is required
to maintain the reaction temperature around 20-30o
C the mole ratio of formaldehydeto acetaldehyde 5:1.A ratio of 1:1.5 mole hydroxide ion per mole of acetaldehyde
appears to the optimum amount of condensing agent. The crude reaction mixture is
then transferred to neutralization reactor. Formic acid is added here to reduce the Ph
of the solution to 7.8-8.0 & to remove sodium ions present as Sodiumformate.The
solution is then evaporated to a specific gravity of 1.270.ITt is then chilled to
crystallize Pentaerithritol and resulting slurry is centrifuged & dried to obtain
crystalline Pentaerithritol as final product.The mother liquor goes to recovery system to recover Sodiumformate by
evaporative crystallization followed by centrifuging & drying to obtain crystalline
Sodiumfomate as by product.
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3.1.5 Chemical Properties of Pentaerithritol
1) Molecular Formula - (CH2OH)4C
2) Molecular Weight - 136.15
3) Boiling Point - 276C AT 760 mm of Hg
4) Melting Point - 262C
5) Density - 1.1318gm/cc
6) CAS No. - 115-77-5
3.1.6 Physical Properties of Pentaerithritol
Pentaerithritol are crystalline solids.
Those solids are soluble in water.
Solubility decreases with increase in molecular weight.
Their solubility and melting points show alternation or oscillation from one
member to the near.
It is moderately soluble in cold water, quite soluble in hot water, and has only
limited solubility in organic liquids.
It is an optically inactive compound, resembling cane sugar in appearance
and has a sweet taste characteristic of polyols.
Pentaerithritol is nonhygroscopic and stable in air and sublimes slowly on
heating.
3.1.7 Process flow diagram
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Figure 3.1: Process flow diagram (Manufacturing of pentaerithritol)
3.1.7 Process flow diagram
TO VACCUM
HCHO
NaOH REACTOR HCOOHCH3CHO NEUTRALISER EVAPORATOR
C.W OUT
CW IN
TO VACCUM
AIR OUT
P.E.T
CRYSTAL
HOT AIR IN HCOONa + water
HOLDIN
G
TANK
VACUM
CRSTALLI
SER HOLDI
NG
TANK
CENTRIFUGE
DRIER
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TO VACUM
VACCUM CRSTALLISER
TO ATM
DRIED HCOONa
MOTHER LIQUOUR
HOT AIR IN
CENTRIFUGE
DRIER
CENTRIFUGE
ML(P.E.T)
Figure 3.2: Process flow diagram (Recovery of sodiumformate)
3.2 MATERIAL BALANCE
3.2.1 Material balance for P.E.T and sodiumformate manufacturing:
BASIS:600 tons/annum
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2000 kg/day of Pentaerithritol
REACTION:
4 HCHO + CH3CHO + NaoH (CH2OH) 4C + HCOONa
(4 x 30) 44 40 136.15 68
3.2.2 Material balance around Reactor:
Reactor input
CH3CHO required = kgmoleCH3CHO/kgmole (CH2OH) 4C x kg P.E.T
= (44/136.15) / 2000kg of P.E.T
= 646.34 kg/day
For 85% conversion = CH3CHO kg required / 0.85
= 646.34 / 0.85= 765 kg/day
CH3CHO purity(99%) = 765 / 0.99
= 757.5 kg /day
HCHO required = (CH3CHOkg/kgmoleCH3CHO) x kgmoleHCHO
= (757.5 / 44) x 30
= 2065 kg/day
Mole ratio CH3CHO: HCHO= 1: 4.55
= (CH3CHOkg/kgmoleCH3CHO) x
kgmoleHCHOx4.55
= 4.55 x (757.5 X 44) x 30
= 2350 kg/days
HCHO purity (37%) = 2350/0.37
= 6350 kg/day
NaoH required = (CH3CHOkg/kgmoleCH3CHO) x kgmole NaOH
= (757.5 / 44) x 40
= 688.4 kg
Mole ratio CH3CHO:NaOH = 1 : 1.52
= kg CH3CHO x kgmoleNaOH x 1.52
= (757.5) x 40 x 1.52
= 1050 kg
NaOH purity(50%) = 1050 / 50
= 2100 kg/day
Reactor output:
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P.E.T produced = (CH3CHOkg/kgmoleCH3CHO)xkgmoleP.E.T
(757.5/44) x136.5
= 2341.23 kg/day
HCOONa produced = (CH3CHOkg/kgmoleCH3CHO) x kgmoleHCOONa
= (757.5 /44) x 68= 1170.0 kg/day
Material balance around reactor:
Component Input
(kg/day)
Output
(kg/day)
CH3CHO 757.50 -HCHO 2350.00 284.70
NaoH 1050.00 361.60
(CH2OH) 4C - 2341.23
HCOONa - 1700.00
H2O 5057.50 5057.50
TOTAL 9125.00 9125.00
3.2.3 Material balance around Neutralizer:
NaOH + HCOOH HCOONa + H2O
40 46 68 18
Neutralizer input
NaOH Unconverted = 316.6kg
HCHO required for neutralization = (kgmoleHCHO/kgmoleNaOH) x 316.6
= (46/40) x 316.6= 415.84 kg
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HCHO purity (34.6%) = 415.84 / 0.346
= 1200 kg/day
Neutralizer output:
HCOONa produced = (kgmoleHCOONa/kgmoleNaOH) x 316.6
+1170kg= 613.94 kg + 1170 kg
= 1783.94 kg/day
H2O produced = kgmoleH2O/kgmoleNaOH x 316.6
= (18 / 40) x 361.6
= 946.88 kg
= 5057.5 +946.88
= 6005.5 kg/dayMaterial balance around Neutralizer:
3.2.4 Material balance around
Evaporator
Data required:Out going slurry:
Material Input
(kg/day)
Output
(kg/day)
P.E.T 2341.23 2341.23
HCOONa 1170.00 1783.94
NaOH 361.60 ****
HCHO 284.70 284.67
HCOOH 1200.00 ****
H2O 5057.50 6005.50
TOTAL 10415.00 10415.00
Component Specific gravity Weight(kg) Volume(lt)
H2O 1.0 (x) (x)
P.E.T 1.396 2341.23 1677.10
HCOONa 1.8 1783.94 991.08
TOTAL ***** 4125.17 + x 2668.18+ x
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Specific gravity of out going slurry = 1.27
(Weight / volume) of out going slurry = 1.27
(4125.17 + x) / (2668.18 + x) = 1.27
H2O along with out going slurry (x) = 2727.94 kg/day
H2O evaporated = 6005.16 2727.94= 3277.22 kg/day
Material balance around Evaporator
3.2.5 Material balance around Vacuum crystallizer:Assume -14.8% loss in solution
P.E.T as crystal = 2341.23 x 0.148
= 341.23 kg/day
= 2341.23 341.23
= 1194. kg/day
Material balance around Vacuum crystallizer
Material Input
(kg/day)
Output
(kg/day)
P.E.T 2341.23 2341.23
HCOONa 1783.94 1783.94HCHO 284.67 ****
HCHO(evaporated) **** 284.67
H2O(evaporated) **** 3272.22
H2O(along slurry) 6005.50 2727.94
TOTAL 10415.00 10415.00
MATERIAL INPUT OUTPUT
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3.2.6 Material balance around Centrifuge:
Assume moisture accompanying with crystal = 9.0 %
Hence moisture with crystal 0.09 = x / (x+1194)
x = 127.28 kg
Material balance around Centrifuge
(kg/day) (kg/day)
P.E.T 2341.23 ****
HCOONa(in solution) 1783.94 1873.94
P.E.T( In solution) **** 347.23
P.E.T(As crystal) **** 1994.00
H2O 2727.94 2727.94
TOTAL 6583.11 6583.11
Material INPUT
(kg/day)
OUTPUT
(kg/day)
P.E.T 1194.0 1194.0
H2O 127.8 6.0
H2O(to ***** 121.8
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3.2.7 Material balance around Dryer:
Centrifuge mother liquor containing HCOONa + Water is sent to Evaporative vacuum
crystallizer for HCOONa recovery
atm)
TOTAL 2121.8 2121.8
Material INPUT
(kg/day)
OUTPUT
(kg/day)
P.E.T 1994.00 *****
HCOONa (in solution) 1783.94 1783.94
H2O( in solution) 2727.94 2600.66
P.E.T(As soluble) 347.23 347.23
P.E.T (in wet cake) ***** 1994.00
H2O (in wet cake) ***** 127.28
TOTAL 6853.11 6853.11P.E.T PRODUCED2000 kg/day
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3.2.8 Material balance around Evaporative vacuum crystallizer:
Data required:
Out going slurry:
Specific gravity of out going slurry = 1.37
(Weight / volume) of out going slurry = 1.37
(2131.17 + x) /(1239.81 + x) = 1.37H2O along with out going slurry (x) = 1185.0 kg / day
H2O evaporated = 2600.66 1185.0
= 1415.66kg/day
Material balance around Evaporative vacuum crystallizer
3.2.9 Material balance around Centrifuge:
Assume moisture accompanying with crystal = 4.0 %
Hence moisture with crystal 0.04 = x / (x+1783.94)
x = 71.1 kg
Component Specific gravity Weight(kg/day) Volume(lt)
H2O 1.0 (x) (x)
P.E.T 1.396 347.23 248.73
HCOONa 1.8 1783.94 991.08
TOTAL ***** 2131.17 + (x) 1239.81 + (x)
COMPONENT INPUT
(kg/day)
OUTPUT
(kg/day)
P.E.T 347.23 347.23
HCOONa 1783.94 1783.94
H2O 2600.00
H2O evaporated ***** 1415.66
TOTAL 4731.83 4731.83
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3.2.10 Material balance around Dryer:
3.3 ENERGY BALANCE
3.3.1 Energyrgy balance around Reactor:
Mass of Acetadehyde(M CH3CHO) = 757.50 kg
Mass of Sodium hydroxide (MNaoH) = 1050.00 kg
Mass of Formaldehyde (M HCHO) = 2350..00
Mass of Sodiumformate(M HCOONa) = 1170.00 kg
Mass of P.E.T(MP.E.T) = 2341.23 kg
COMPONENT INPUT
(kg/day)
OUTPUT
(kg/day)
HCOONa 1783.94 *****
HCOONa (in ML) ***** 89.94
H2O(in ML) 1185.00 1113.9
P.E.T(in ML) 347.23 347.23
HCOONa (in wet cake) ***** 1694.00
H2O (in wet cake) ***** 71.00
TOTAL 6853.11 6853.11
COMPONENT INPUT
(kg/day)
OUTPUT
(kg/day)
HCOONa 1694.00 1694.00
H2O 71.00 6.00
H2O(to atm) ***** 65.0
TOTAL 1765.00 1765.00
HCOONa PRODUCED1700 kg/day
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Feed inlet temperature = 25C
Product outlet temperature = 30C
Reaction temperature = 80C
Reference temperature = 25C
Energy balance equation
Heat input + (MP.E.T x heat of reaction) =Heat output+Heat load on Reactor
Cooling jacket
Heat input
Feed temperature = reference temperature = 25C
So, heat input = 0
Table 3.1 Heat Values of Components
COMPONENTREACTANT PRODUCT
HCHO NaOH CH3CHO HCOONa P.E.T
Specific heat
capacity
(Kcal/gmoC)
0.526 0.540 0.312 0.269 0.386
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Heat of
formation
(kcal/gmole)
-60.57 -72.01-142.53
-226.0 -100.0
Heat of reactionat 25 oC = HF product - HF reactant
= (-326.65) - (-275.11)
= - 51.49 Kcal/gm mole
Enthalpy at 25 oC = (-51.49x 2341.231000)/136.15
= -886396.56 Kcal/day
HF reactant = m.Cp.dt
= ((2350 x 0.526)+(1050x0.54)+
(757.5x0.312)(298-303)
= -10199.1 Kcal/day
HF product = m.Cp.dt
= (2341.23 x 0.386) + (1170 x 0.269) x 5
= 6092.2 Kcal/day
Q = HF product+HF reactant+ heat of reaction at 25oC
= -101991.1 + 6092.2 - 886396.56
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Q = -890503.46Kcal/day
Heat loss by
radiation 10% = -89050.35Kcal/day
Heat to be removed
(Q) = 890503.46 - 89050.35
= 801453.117Kcal/day
Heat removed = Heat generated = Q = (mCpdt)Water
Q = m*1*(303-293)
= 801453.11 Kcal/day
Mass of H2O
required ( m ) = 801453.11 / 10
= 80145.31 kg/day
3.3.2 Energyrgy balance around Holding Tank:
Here the reactor outlet mass is heated up to 90oC before feeding to evaporator.
Temperature of the
holding tank = 90 oC
Heat to be supplied = mCpdt
= (2341.23 x 0.386 +1783.94 x 0.269 + 6005.16 x 1+
284.67x .576) x (90-30)
= 453154.75 kcal/day
Heat loss due to radiation = 10%
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Haet required = Heat supplied
= (10%)x453154.75
= 1.1*453154.75
= 498470.2338kcal/day
Heat loss by radiation = 498470.23-453154.75
= 45315.47kcal/day
Steam required @ 2.1125
atm Ms = 498470.23/525.
= 948.3kg/day
Heat loss by condensation = 121.27x948.3
= 115000 kcal/day
Heat to be supplied by
steam = (525.64-121.27) x 948.3
= 383471.19kcal/day
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3.3.3 Energy balance around dryer:
Amount of P.E.T to be dsried = 1994 kg/day
Water to be removed = 121.28kg/day
Sensible heat required to heat material from 25o
C -100o
C =mCpdt
= (1994\x (100-75) x (0.386+ 121.80) x
(100-75) x 1
= 66822.3 Kcal /day
Heat required for vaporization of 121.28 kg of water
= 121.28 x 540
= 65491.2 Kcal/day
Hence total heat required = 66822.3+65491.2
= 132313.50kcal/day
Assuming 10% heat loss due to radiation
Hence heat required = 1.1 x 132313.50
= 145544.85 kcal/day
Steam required = 145544.85/525.64
= 276.88kg/day
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3.3.4 Energy balance around Evaporator:
BASIS:
P.E.T-2000kg/ day.
Data required for calculation:
As per evaporator material balance,
Feed = 10415.00 kg/dayVapour = 3561.89kg/day
Product = 6853.11kg/day
Feed enters at = 90oC
Steam table data
Steam,2 atm,121oC (HS) = 645.26 Kcal/kg
Condensate,1 atm,105oC(HC) = 105.60 Kcal/kg
Table 5.1 Heat Values of Components
General heat balance equation for evaporator
F.HF + S(HS-HC) = E.HE + PHP 1
REACTANT PRODUCT
HCHO NaOH CH3CHO HCOONa P.E.T
Specific heat
capacity
(Kcal/gmoC)
0.526 0.540 0.312 0.269 0.386
Heat of
formation
(kcal/gmole)
-60.57 -72.01-142.53
-226.0 -100.0
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Feed (F) = 10451 kg/day
Enthalpy of feed = (mCpdt) feed
= (2341.23x0.396) + (1783.94x0.269) +
(6005.5x1) x 90
F. HF = 679708.08 kcal/day= 28321.17 kcal/hr
Vapour (E) = 3561.89 kg/day
= Water vapor evaporated x Enthalpy at 100
oC
= 3561.89 kg/day x 644.22 kcal/kg
= 2294640.776 kcal/day
= 95610.03 kcal/day
Product (P) = 6853.11kg/day
Enthalpy of product (HP) = (mCpdt) product
= (2341 x 0.368) + (2727.94 x1)
+ (1783.94 x 0.269) x 90
= 411152.4kcal/day
= 17131.39kcal/day
Substituting the above values in equation -1
Mass of steam(S) = 3717.20kg/day
Steam economy = (kg water evaporated) / (kg of
steam required)
= (3561.89/3717.2)x10
= 0.95
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3.4 EQUIPMENT DESIGN
3.4.1 Design of vacuum tray drier
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Figure 3.1: Design of vacuum tray drier
BASIS :
48 Tray capacityShell :
Length of the shell L = 1.7m
Breadth of the shell b = 0.91m
Area of the shell = Lx B
Area of the shell = 1.547 m2
Tray:
Length of the tray L = 0.8mBreadth of the tray b = 0.4m
Area of the tray = Lx B
Total number of tray = 48 number
Area of the tray = 0.32m
Total area of the tray = 48 x 0.32
= 15.36 m2
Shelves:
Length of the shelves L = 1.2m
Breadth of the shelves b = 0.8m
= Lx B
= 16
Area of the shelves = 0.96m
Total area of the tray = 16 x 0.96
Overall surface area of the drier = (1.547+15.6+15.36)
= 32.27m2
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3.4.2 Design of Evaporator
Circulation and heat transfer in this type of evaporator is strongly affected by
the liquid level as indicated by an external gauge, which is only about half way of the
tube. Slight reduction in level below the optimum results in incomplete wetting of the
tube walls with a consequent increase in tendency to foul and causes rapid reductionin capacity. When this type of evaporator is used with a liquid that can deposit scale,
it is customary to operate with the liquid level appreciably higher than the optimum
level which is above the top tube sheet.
Advantages of Short Tube Vertical Evaporator:
High heat transfer coefficient
Low head
Easy mechanical descalingMild scaling solution can be used for mechanical cleaning as the tubes are short
and large in diameter.
Crystalline products can be removed using the agitatorProcedure followed in design
of evaporator assembly (ref: Unit operation of chemical engineering Mc Cabe)
Q = U.A dt
Q = 1316857.14
Q = 1316857.143 x 0.252 = 331848 Btu/hr
U = 160 Btu/hr.ft2. oF
Ref: Unit operation of chemical
JM Coulson & JF RICHARDSON PAGE-520
T = 45o
F
Hence A = (331848) / 45 X 160) = 46.09 ft2
For Sheet Diameter:
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C/S Area of one tube = /4) d2)
= (3.14/4) x (2.2)2x48
= 182.463 in 2
From Mc Cabe, 30% of A is the area of downcomer
= 182.463 x 0.3
= 54.4 in2
( /4) d2 = 54.4 in2
Hence, d = 8 inches
Length and Number of tubes:
In standard short tube evaporators the length varies from 4 to 8 ft and the
diameter is around 2 to 4 inches, In this case, the length is assumed as 4 ft and
diameter as 2 inches.
Outer area of each tube = d0l
= 3.14 X (2.12/12) X4 = 2.1ft2
Number of tubes required = 46.09/2.1
= 22.2 tubes
Hence we can take 22 to 24 tubes.
A staggered arrangement is used as it permits higher tube accommodation, for a
given distance between the tubes. From the approximate calculations, the shell
inside diameter is taken as 21.2 inches.
Thickness of Shell = (P.d) / (2.f.E) + Corrosion Allowance
= (30 X 21.2) (2X0.8 X 3312.5) + 0.08
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= 0.2 inches maximum
Hence Outer Diameter = 21.2 + 0.2 + 0.2
= 21.6 inches
Length of the evaporator is proportioned with respect to the length of tubes.
Design of evaporator and its accessories:
The standard evaporator consists of a vertical cylinder with calendria cross
which the heat exchange takes place. The cylindrical body terminates at the top in a
save all , the objective of which is to separate the liquid droplets which maybe
entrained with vapour from the soilution. Previously the evaporator body was
fabricated with cast iron; however more recently fabrication using steel plate is
becoming more common.
Height of the Vessel:
The space above the tubular calendris represents the greater part of the volume
taken up by the equipment. The objective is to diminish the risk of entrainment
droplets of solution projected by boiling. Various MOC used as follows:
Part Old Modern Special
Shell Steel Bronze Mild Steel Stainless Steel
Tubes Brass, Cast Iron Mild Steel -
The height of the cylindrical portion above the steel plates is 1.5 to 2 times the length
of the tube.
Calendria:
It is the continuation of the shell or body of the evaporator. It is often fixed to
the shell. The bore of the holes provided in the tube plate is about 1/32 inch greater
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than the outer diameter of the tubes. Vertical baffles are often placed in the calendria
with the object of compelling the steam to follow a certain path.
Center Wall (Downcomer):
The calendria is generally designed with a wide tube or center wall. Solution
which has been projected over the top tube plate is returned to the bottom by the
downcomer. This center well is often used to collect the concentrated solution in
order to transfer it from vessel to other.
Air source for condenser:
Air introduced into the condensers comes from various sources such as air
contained in the heating system, air introduced in the cold rejection water, air
entering by leakage.
Specification sheet of evaporator
Number Required 1
Type Short tube, vertical calendria
type evaporator
Normal Capacity 328.25 l/hr
Working Pressure 2 atm
Critical Dimension Overall height = 220 inch
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Inside diameter of Shell = 21.2 inch
Inside diameter of tube = 2 inch
Thickness of shell = 0.2 inch
Thickness of tube = 0.1 inch
Diameter of downcomer = 3 inch
Length of tube = 48 inch
Number of tubes = 22 to 24
Material of construction Stainless steel
Baffles As per requirement
Nozzles or jet stream As steam inlet to increase velocity
Cost Rs.6,00,000 (Approx)
CHAPTER 4
COST ESTIMATION,PLANT LATOUT,MATERIAL SAFETY
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4.1 COST ESTIMATION
4.1.1 Estimation of Direct and Indirect cost:
Marshall and Shift index in 1992 for equipments is 943.1
Cost estimation based on the equipment
Cost of the reactor = Rs 1.7 x 106
Cost of the Neutraliser = Rs 0.62x106
Cost of the Holding tank = Rs 1.1x106
Cost of Crystallizer = Rs 0.71x106
Cost of pump = Rs 0.25x106
Cost of the Dryer = Rs 0.25x106
Cost of the Centrifuge = Rs 0.32x106
Total cost of the equipment = Rs 4.3x106.
4.1.2 Estimation of total direct cost:
Purchased equipment cost ( E ) = Rs 4.3x106
Purchased equipment installation (39% of E) = 4.3x106x 0.39
= Rs 1.677x106
Instrumentation (installed cost), 28%E = 4.3x106x0.28
= Rs 1.204x106
Piping installed, 31%E = 4.3x106x0.31
= Rs 1.333x106
Electrical installation, 10%E = 4.3x106x0.1
= Rs 0.43x106
Yard improvement, 10%E = 4.3x106x0.1
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= Rs 0.43x106
Service facility, 55%E = 4.3x106x0.55
= Rs 2.365x106
Land, 6%E = 4.3x106x0.06
= Rs 0.258x106
Total direct cost (D) = Rs 11.997x106
4.1.3 Estimation of total indirect cost:
1. Engg and supervision ( 32% E ) = Rs 1.376x106
2. Construction + contractor fees (25% direct costs)
= 11.997x106x0.25
= Rs 2.99925 x106
Therefore total indirect cost (I) = Rs 4.37525x106
DIRECT AND INDIRECT COST (TOTAL) = Rs 16.37225x106
Contingence (10%D+I) = (4.37525+0.1x11.997x106)
= Rs 5.57495x106
Fixed capital investment (FCI)
Contingence + D + I = Rs 21.9472x106
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Working Capital: (10-20% of Fixed-capital investment)
Consider the Working Capital = 15% of
Fixed-capital investment
i.e., Working capital = 15% of 21.9472*106 = 0.15 21.9472x106
= Rs 3.29208x106
Total Capital Investment (TCI):
Total capital investment = Fixed capital investment +
Working capital
= Rs 25.23925x106
i.e., Total capital investment = Rs 25.23925x106
4.1.4 Estimation of Total Production cost:
Manufacturing Cost = Direct production cost +
Fixed charges +
Plant overhead cost.
4.1.5 Fixed charges:
(10-20% total product cost)
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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 = Rs. 2.85313106
ii. Local Taxes:(1-4% of fixed capital investment)
Consider the local taxes = 3% of fixed capital investment
i.e. Local Taxes = 0.0321.9472*10
6
= Rs. 0.6584106
iii. Insurances:(0.4-1% of fixed capital investment)
Consider the Insurance = 0.7% of fixed capital investment
i.e. Insurance = 0.007s Rs 21.9472*106
= Rs. 0.15363106
iv. Rent: (8-12% of value of rented land and buildings)
Consider rent = 10% of value of rented
land and buildings
Rent = Rs. 0.13023x106
Thus, Fixed Charges = Rs. 3.79539106
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4.1.6 Direct production cost:
(about 60% of total product cost)
Now we have Fixed charges = 10-20% of total product charges
Consider the Fixed charges = 15% of total product cost
Total product charge = fixed charges/15%
= 3. 7953910
6
/15%
= 3. 79539106/0.15
Total product charge (TPC) = Rs. 25.3026106
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 25.3026106
= Rs. 6.32565106
ii. Operating Labour (OL):(10-20% of total product cost)
Consider the cost of operating labour = 12% of total product cost
Operating labour cost = 12% of 25.3026106
= Rs 3.0363106
iii. Direct Supervisory and Clerical Labour (DS & CL):
(10-25% of OL)
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Consider the cost for Direct
supervisory and clerical labour = 12% of OL
Direct supervisory and clerical
labour cost = 12% of 3.0363
= Rs. 0.364357106
iv. Utilities: (10-20% of total product cost)
Consider the cost of Utilities = 12% of total product cos
t
Utilities cost = 12% of 25.3026106
= Rs. 0.12 25.3026106
= Rs. 3.036312106
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 = Rs. 0.0525.23925*106
= Rs. 1.26196106
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 1.26196106
= Rs. 0.18929106
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vii. Laboratory Charges: (10-20% of OL)
Consider the Laboratory charges = 15% of OL
= 15% of 3.0363106
Laboratory charges = Rs. 0.45544106
viii. Patent and Royalties:(0-6% of total product cost)Consider the cost of Patent and
royalties = 4% of total product co
st
= 4% of 25.3026106
Patent and Royalties cost = Rs. 1.0121106
Thus, Direct Production Cost = Rs. 15.68141106
4.1.7 Plant over head cost:
(50-70% of Operating labour, supervision, and maintenance or 5-15% of total
product cost); includes for the following:
general plant upkeep and overhead, 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 ((3.0363106)
+ (0.364357106)
+ (1.26196106)
Plant overhead cost = Rs. 2.79765106
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Thus, Manufacture cost = Direct production cost +
Fixed charges +
Plant overhead costs.
Manufacture cost = 15.68141106 + 3.79539106
+ 2.79765106
= Rs 22.27445106
II. General Expenses = Administrative costs+ distribution and selling costs
+ Research and development costs
Administrative costs :( 2-6% of total product cost)
Consider the Administrative costs = 5% of total product cost
Administrative costs = Rs. 1.26513106
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 25.3026106
Distribution and Selling costs = Rs. 3.75453106
Research and Development costs: (about 5% of total product cost)
Consider the Research and
development costs = 5% of total product cost
= 5% of 25.3026106
Research and Development costs = Rs. 1.26513106
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Financing (interest):(0-10% of total capital investment)
Consider interest = 5% of total capital investment
i.e. interest = 5% of 25.23925x106
Interest = Rs. 1.26196106
Thus, General Expenses = (3.75453+1.26515+1.26195)
x106
= Rs 6.29163x10
6
III. Total Product cost = Manufacture cost +
General Expenses
Total product cost = 22.27445106+ 6.29163x106
= Rs 28.55608 x106
Gross Earnings/Income:
Wholesale Selling Price of P.E.T per ton = $ 650
Hence
Wholesale Selling Price of P.E.T per ton. = 650 46
= Rs. 29900
Total Income = Selling price Quantity of
product manufactured
= 29900 (2 T/day)
(300 days/year)
Total Income = Rs. 17.94 106
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Gross income = Total Income Total
Product Cost
= Rs 17.94106 28.556106
Gross Income = Rs10.62106
Let the Tax rate be 45% (common)
Taxes = 45% of Gross income
= 45% of 10.62106
Taxes = Rs. 4.61106
Net Profit = Gross income Taxes
= Gross income
(1- Tax rate)
Net profit = (10.62106) (4.62106)
= Rs. 6.01106
4.1.8 Rate of return:
Rate of return = (Net profit100)/Total Capital Investment
Rate of Return = (6.01106100)/ (25.23925106)
Rate of Return = 23.99%
4.1.9 Payback period:
Payback period = (Total capital investment+total product cost) /net profit
= (25.239 x 106+28.556x106 )/ 6.01x106
= 4.3954 years
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4.2 INSTRUMENTATION
The primary objective of a designer when specifying instrumentation and control.
Safe plant operation
To keep the process variable within known safe operating limits
To detect dangerous situations as they develop and to provide alarmsautomatic shutdown systems
To provide interlocks to prevent dangerous operating procedures
To achieves the desired product output
To maintain the product composition within the specified quality standards.
Process may be controlled more precisely to give more uniform and higher
quality products by the application of automatic co troll often leading to higher profitautomatic is also beneficial in certain remote, hazardous or routine operations
For manual control, an operator periodically measures the temperature .if it is below
the desired value, the operator increases the steam flow by opening the valve
slightly for automatic control ,a temperature sensitive device is used to produce a
signal (electrical , pneumatic etc.) propositional to the measured temperature .
This signal is fed to a controller, which compares it with a preset desired
values or set point . If a difference exists, the controller changes the opening of the
steam control valve to correct the temperature.
4.2.1 Control valve:
The control valve contains a pneumatic device that moves the valve stem as
the pressure on a spring loaded diaphragm changes .the stem positions a plug in the
orifice of the valve body. as the pressure increases ,the plug moves downward and
restricts the flow of fluid through the valve.
The valve may also be constructed to have air to- open action. Most
commercial valves move from fully open to fully closed as the valve top pressure
changes from 3to 15psi. In general the flow rate of fluid through the valve depends
upon the upstream and downstream fluid pressure and the size of the opening
through the valve
4.2.2 Controler:
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The control hardware is required to control the temperature of a stream
leaving the heat exchanger. In general a controller contains
Transducers (temperature to current )
Controller-recorder (current to current)
Converter(current to pressure) Pressure valve (pressure to flow rate)
Thermocouple is used to measure the temperature, the signal from the
thermocouple is sent to the transducer, which produces an output in the of 4-20mA,
which is a linear function of the input. The output of the transducer enters the
controller where it is compared to the set point to produce an error signal. The
controller converts the error to an output in the range of 3-16 psig which is a linear
function of the input.Finally, the output of the converter is sent to the top of the control valve,
which adjusts the flow of cooling water to the heat exchanger. Electricity is needed
for the transducer, controller and converter. A source of 20psig air is needed for the
converter.
In a well-tuned control system, the response of the temperature will oscillate
around the set point before coming to a steady state.
4.2.3 Instrment In Batch Reactor:
Pressure gauge,
Temperature indicator,
Steam control valves,
Cooling water control valves,
Radar level indicator,
Differential pressure level indicator,
Pressure transmitter,
Orifice plate,
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4.3 PLANT LOCATION AND LAYOUT
4.3.1 Plant Layout and Site Selection:
The following are the three necessary for the general plant layout
The starting point or location or reference A kinetic diagram or directional factor
A statement of the special requirement for various product ,raw material ,by
product storage facilities and for processing department
From known calculated or estimated space requirement for individual equipment.
It is possible to obtain proper allocation of equipment and overall dimensions of the
building
Process Plant PipingThe American national standard institute (ANSI) and the American petroleum
institute (API) have established dimensional standard for the most widely used piping
components they list these standard of the American society for testing material
(ASTM) and American welding society(AWS).The consideration to be evaluated
when selecting pipe material are:
Possible exposure to fire respect to loss of strength, degradation
Ability of thermal insulation
Susceptibility of the pipe to brittle failure
Susceptibility of piping material to crevice corrosion
The Susceptibility of packing ,seal gasket as well as compatibility with the fluid
handled
The refrigerating effect sudden loss of pressure as volatile fluid in determining
the
lowest expected service temperature
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The organization which generate standards of major importance to chemical
process industries in United States are tested
American national standard institute (ANSI)
American society of mechanical engineers(ASME)
American society for testing material (ASTM) International organization for testing material (ISO)
The economic consideration and efficient operation of a process unit depend on
how well plant and equipment specified on the process flow sheet is laid out .the
major factors to be considered are
Economic consideration construction and operating cost
Damage to persons and property in case of fire explosion and toxic release
Future expansion Modular construction
In preparing the layout, one has to satisfy various regulation acts which are in
force in the area where the plant is located. The main considerations are
Factories act
The explosives act
Excise rules
Health rules
The boiler act
4.3.2 Plant Layout
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Figure 4.1 Plant Layout
4.4 MATERIAL HANDLNG AND SAFETY
Safety is one of the important aspect in any chemical industry. To run the
process effectively, safety rules and regulations has to be followed. the guide lines
are provided by the government and the industries has to implement them. Here we
present about the safety guidelines adopted in ammonia plant
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In any industry, whether it is a small or large scale industry, safety must be
emphasized .there is a safety department to look after the safety procedures inside
the industry
4.4.1 Kinds of Safety:
Careful precautions are given to the workers upon their work
The safety officer looks after the loading and unloading operating operations
inside the plant
There is an emergency control center for meeting in the emergency period
There are assembly points inside the plant for the assembling of the workers
during the emergency period.
Mock drills are conducted in every 3months and they check with theiravailabilities
The management is conducting seminars for the workers once in a month on the
related areas such as safety, occupational health hazard, first aid, etc.
PPEs of all kind are provided to the workers
Fresh air line and showers are provided
Medical center is established for emergency situations
4.4.2 Safety Equipment:
Personal safety
Equipment safety
Material safety
Personal Safety is for personnel proteciton
Respiratory
Non respiratory
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Respiratory is used for to ovid inhalation of chemical
Air supplying respiratory
Air purifying respiratory
Other safety protection is mentioned below.
Helmet Safety shoe
Gloves
Apron
Goggles, etc .
Equipment Safety is to safe operation. Some of the safety feature
adopted in process is listed below. Rupture disc
Pressure safety valve
Set points
Alarm
Emergency shut off
Trips
Material Safety
Set points
Controls to temperature, pressure, ph, etc
Standard operating procedure
Material safety data sheet
4.4.3 Fire Fighting Equipment:
Mechanical form
DCP etc,
Hydrant
Single headed fire hydrant
Double headed fire hydrant
Fire water monitor
fire escape hydrant
Foam trally With water monitor
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4.4.4 Safety in Batch Reactor:
Rupture disk
pressure safety valve
alarm system
tripping system set points
Interlocks & control
A rupture disc or bursting disc is a pressure relief device that protects a
vessels or system from over pressurization.
Rupture discs have a one-time use thin foil that fails at a predetermined
pressure, either positive or vacuum.
Rupture discs provide fast response to an increase in system pressurebut once the membrane has failed it will not reseal.The equipment must be
protected against being subjected to an internal vacuum (i.e., low
pressure) that is lower than the equipment can withstand.
Vacuum relief valves are used to open at a predetermined low pressure
limit and to admit air or an inert gas in to equipment so as control the amount
of vacuum.
If the chemical is drawn from the tank additional nitrogen pressure is fed to
the tank by the blanketing regulator system. In case, if the nitrogen is not
available from the source, then the tank may go under negative pressure and
vacuum is created in the tank. due to this the tank may collapse. to avoid such
accidents vacuum relief valve breathes in atmospheric air and protects the
tank from further damage.
Interlock is a device used to help prevent a machine from harming its
operator or damaging itself by stopping the machine when tripped.
Safety protection requirements are interlocked with the equipment
operation to save the plant and person. Also called permissive.
Interlock testing will be done based on the inter lock report which was
already tested and certified. Interlock testing duration intervals are defined
based on the criticality of the equipment.
Is the condition to start or stop the equipment or valve or pump etc.
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If the interlocks are satisfied than only you can able to operate the require
operation.
For example if the centrifuge lid is in open condition you can not able to
start the centrifuge.
Interlock check reports are available for all critical equipments and testedperiodically.
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The following controls are used in batch reactor
The flow rate of steam can be controlled by the manual set point of
percentage of valve opening.
The flow rate of steam can be controlled by setting the requirement ofsteam flow rate. Based on the steam flow rate the steam control valve will
be regulated automatically.
Manual mode:
The flow rate of cooling water to the condenser can be controlled by the
manual set point of percentage of valve opening.
Auto mode:
The flow rate of cooling water to the condenser can be controlled by settingthe requirement of cooling water flow rate based on the cooling water flow
rate the control valve will be operated automatically.
Cascade mode:
The flow rate of cooling water to the condenser can be controlled based on
the heat load of condenser.
Temperature control:
The flow rate of cooling water can be controlled based on the cooling water
out let temperature of total condenser.
4.5 EFFLUENT TREATMENT PLANT
Effluent is any water that has been adversely affected in quality by pollutants
and the same drained during the process of production by production and other
service department
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4.5.1 Effluent origin:
Utilities cooling waters(biocides ,heat ,slimes ,silt)
Production process waste water
Production process waste aqueous solution Residue from process
Above Effluent Contains the Following
High COD
Extreme pH-from Process
Traces of solvent
4.5.2 Need for Effluent Treatment: Treated effluent should comply with the local and/or national regulation
regarding disposal of effluent into community treatment plants or into rivers
,lakes or ocean
For the above orchid established zero discharge effluent treatment facility
so that the pollutant concentrations in the effluent is treated and reused for
the utilities (water make up for cooling towers, in recovery column water
reflux and solvent washing ,in ETP filtration system water washing)
Effluent Treatment
It is a group of unit process designed to separate ,modify remove ,and
destroy undesirable substance carried by effluent sources
Effluent Quality Indicator
The quality of the effluent is characterized by the following parameters
pH
COD (hemical oxygen demand )
BOD (Biological oxygen demand )
DO(Dissolved oxygen)
TDS (Total dissolved solids)
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Others (silica, Iron, Fluoride)
4.5.3 Types of Effluent:
Based on the incoming effluent characteristics we separated effluent into two
streams for effective treatment to achieve the zero dischargeThe two effluent streams are
1. LPS-Low polluted stream
2. HPS-High polluted stream
LPS &HPS Standard are mentioned as belove
LPS
1. COD Less than 15,000ppm
2. fluoride ,iron,& traces solvent nilHPS
1. COD more than 15,000ppm
2. traces of Solvent
LPS and HPS are treated as belove mentuned
LPS (Low pollutant stream)is the effluent source which is treated by the bio
chemical and filtration process to reuse for utilities
HPS (High pollutant stream)is the effluent source which is treated by stripping
evaporation and distillation processes to convert into LPS source and to reuse
for
cooling towers and other applicable usages
4.5.4 Effluent Treatment Processes:
Treatment stages in LPS
Primary treatment
Secondary treatment
Advanced treatment
Primary treatment
Primary treatment or pretreatment processes address the problem of
screening ,equalization , neutralization, removal of floating scum and removal
of suspended matters
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Screening
Screen grit chamber is used for screening process to separate the solid
materials of more than 6mm from the LPS and to settle heavier particles at
the bottom of the chamber
The above said solid material from the screen and the settled particles are tobe cleaned periodically
By purging air at the bottom of the equalization tank for homogeneous mixing
of LPS equalize the parameters (temperature, etc)and uniform feeding
Any abnormal Ph is observed in the Eeropac(pH is the less than 6.5 and 8.5)
neutralization to be done
The total suspended solids (chemical sludge )which is collected at the center
of the clarifier is taken for circulation from the neutralization tank and thesettled TSS can be thickened further in centrifuge.
Biological method s are cheaper ,effective and eco friendly the following
biological methods are mainly used to reduce COD in the effluent by
maintaining the MLSS&DO
1. Fine bubble jet aeration system
2. Coarse bubble surface aeration system
By fine bubble jet aeration system we are effectively supplying the oxygen
from
the atmosphere air through the air blower to the bottom of the aeropac (circular
tank used for fine bubble aeration)with the effluent emerging from the submersible
pump connected to the FRP kit with spray nozzle .the same set up is kept in the
each compartment of the aeropac.
Coarse Bubble Surface Aeration System
By surface aeration we are effectively supplying the oxygen from the
atmosphere air through the aerator by splashing the surface of the liquid
(rectangular tank used for the surface aeration)
Advance Filtration Treatment Process
This process mainly used to recycle the resource (effluent) and to achieve the
zero discharge.
Ultra filtration system
Nano filtration system
Reverse osmosis system
Ultra filtration system is used for following reason as mentioned below
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After secondary treatment we still have suspended solids in effluent. Using
filtration methods we can trap all the suspended maters.
If the feed water passes through the any filtration system it can be separates
as two components
1. Permeate2. Reject
The permeate water from the UF is taken for NF for further treatment.
The reject water from the UF is recycled in our biological system.
UF filtration steps
Pre filtration
Cartridge filter (S.S) 250 micron pore size
To avoid membrane contamination Fine porous ceramic membrane are used for filtration
.
The Principle Nano filtration system is.Nano filtration working as same as the
osmosis principle only filtration range is differs. When the feed water pass through
the semipermiable membrane (poly amide) the movement of solvent phase (that is
clear filtrate) takes place from the higher concentration to lower with continue till the
applied pressure is greater than the osmotic pressure of the feed water.
Dissolved macro particle shall be removed in this specific range of
The permeate water from the NF is taken for RO for further treatment
The reject water from the NF is send to ecology for HPS treatment
NF filtration steps
Pre filtration steps
Cartridge filter (PP wounded) 5 micron pore size
To avoid membrane contamination
NF membrane filtration
Membrane is made of spirally wounded fine porous poly amide
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Reverse osmosis system
Principle
When the feed water (concentrated)water pass through the semipermiable
membrane (poly amide) the movement of solvent phase (that is clear filtrate) takes
place from the higher concentration to lower with the applied pressure the filtrationwill continue till applied pressure is greater than the osmotic pressure of the feed
water.
Dissolved micro particle shall be removed in this specific range of filtration.
The permeate water from the RO is used for cooling water make up in utilities &
column reflux in the recovery plants.
The reject water from the RO is send to ecology for HPS treatment.
RO filtration steps Pre filtration Cartridge filter (PP wounded) 5 micron pore size
To avoided membrane contamination
RO membrane filtration Membrane is made of spirally wounded fine porous
poly amide.
Treatment process in HPS is mentioned below
Stripping
To remove mixed solvent in HPS
Evaporation
HPS is converted to LPS by evaporation
(Three types of evaporator)
Waste heat recovery
Impure stream heat is transferred into pure stream
Concentration (Concentrated as salt)
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CHAPTER 5
CONCLUSION
A detail study of manufacture of Pentaerithritol has been completed with
material and energy balance calculation for the production capacity of 600 tones per
annum. Design of the reactor and vacuum tray drier involved in the process has
been studied as per requirements & also plant layout, safety, with respect to
environment aspects has been studied. It has been estimated that Payback period
of 4.329 can be achived by suitable selection of process. The effluent treatment for
this process has been studied in detail manner. Safety related to material handling,
fire fighting equipments and safety features of batch reactor is studied in detail.
The application of penaerithritol is studied. By product sodiumformate recoveryis studied in detail with respect to material balance.in overall view it has been
concluded that this method of process adopted for manufacturing of pentarerithritol
and its by- product is most suitable economical process.
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