snk final project

7/29/2019 Snk Final Project

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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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