Manufacturing of Pentaerythritol

MANUFACTURING OF PENTAERYTHRITOL

By

Siddharth MaityB.E., Mumbai University, 2006

An Industrial Engineering Project ReportSubmitted to the Faculty of the

Speed Scientific School of the University of Louisvillein Partial Fulfillment of the Requirements

for the Degree of

Master of Engineeringin

Industrial Engineering

Department of Industrial EngineeringUniversity of Louisville

Louisville, Kentucky

December 2009

MANUFACTURING OF PENTAERYTHRITOL

By

Siddharth MaityB.E., Mumbai University, 2006

An Industrial Engineering Project ReportApproved on

By the Following Faculty Member:

________________________________

Faculty Name

ACKNOWLEDGEMENTS

I take this opportunity to express my sincere gratitude towards Dr. Suraj M. Alexander for his

guidance, help and constant encouragement. This Project Report would not have been possible

without his support.

I also thank Bhageria Dye-Chem for allowing me to visit their plant and providing me with data

required for Material Balance, Energy Balance and Cost Estimation.

ABSTRACT___________________________________________________________________________________________5

INTRODUCTION____________________________________________________________________________________6

Physical and chemical properties__________________________________________________________________7

Reactions of Pentaerythritol_____________________________________________________________________10

Product Applications___________________________________________________________________________13

PROCESS SELECTION_____________________________________________________________________________17

PROCESS DESCRIPTION_________________________________________________________________________19

THERMODYNAMIC FEASIBILITY_______________________________________________________________20

MATERIAL BALANCE_____________________________________________________________________________21

ENERGY BALANCE________________________________________________________________________________29

DESIGN OF EQUIPMENT_________________________________________________________________________34

Design of Neutralization Tank___________________________________________________________________34

Evaporator Design Report_______________________________________________________________________39

LIST OF MAJOR EQUIPMENTS__________________________________________________________________45

INSTRUMENTAL PROCESS CONTROL & PRECAUTIONS_____________________________________50

SELECTION OF LOCATION & PLANT LAYOUT________________________________________________51

SERVICE AND UTILITIES_________________________________________________________________________53

PROFITABILITY & COST OF PROJECT_________________________________________________________54

CONCLUSION______________________________________________________________________________________63

REFERENCES______________________________________________________________________________________64

Abstract

Pentaerythritol, an organic compound has found its use in a wide range of industries due to its

multi-functionality and favorable physical & chemical properties. This project studies the various

chemical and industrial aspects involved in the manufacturing of this compound. Process

selection, Material & Energy Balance, Design of Equipments are some the topics from chemical

engineering which are covered during the course of this project. On the other hand, Plant layout

& selection of location, Service and Utilities, Cost estimation are some topics with industrial

engineering background. This project aims at using and analyzing all these ideas and the

available data to come up with a model which can be used for manufacturing of Pentaerythritol.

Introduction

In the year 1882, pentaerythritol was discovered as byproduct of the reaction between barium

hydroxide and impure formaldehyde. Several year later in 1891, Tollens along with his

coworkers found that the impurity was acetaldehyde, which condensed with the formaldehyde

under mildly alkaline conditions. The name pentaerythritol which was assigned to this compound

was derived from erythritol t o 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. Pentaerythritol 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 a limited solubility in organic liquids.

Uptil 1930 all pentaerythritol produced was used in manufacturing of pentaerythritol tetranitrate,

an explosive substance. Purified pentaerythritol (P.E) is used as a surface coolant. It is also in

paints, varnishes industry and the production of resins. P.E is increasingly used in resins

manufacturing mainly because of its desirable characteristics and price stability.

Physical and chemical properties1. Pentaerythritol

Structure:

CH2OH | |HCOH2--------C-------CH2OH | | CH2OH

Physical form White, crystalline and odourless

Molecular weight 136.15

Composition C = 42.2% H = 8.8% O = 49%

Melting point 262oC

Boiling point 276oC at 30m.m pressure.

Specific gravity 1.396

Toxicity Low and nonirritating to skin

Heat of combustion -660 Kcal/Mole

Heat of formation 226.6 Kcal/Mole

Heat of vaporization 22 Kcal/Mole

Heat of sublimation 31.4 Kcal/Mole

Pentaerythritol sublimes on heating under reduced pressure. It is non hygroscopic, nonvolatile,

optically inactive and stable in air. It is slightly soluble in cold water and readily soluble in hot

water.

2. Sodium Hydroxide (NaOH)

Molecular weight 40

Boiling Point 130.04 K

Melting Point 596.04 K

Liquid Molar Volume 2.245 x 10-2 m3/Kg mol

Heat of Formation -4940.27 KJ/Kg

Gibbs Free energy -5008.47 KJ/Kg

Heat of Fusion 165.13 KJ/Kg

Liquid Cp 2.180 KJ/Kg-K

Density 2101.6 Kg/m3

3. Acetaldehyde (CH3CHO)

Molecular weight 44



Liquid Molar Volume 5.652 x 10-2 m3/Kg mol


Gibbs Free energy -2923.81 KJ/Kg

Heat of Fusion 73.07 KJ/Kg

Heat of Combustion -25054.02 KJ/Kg



4. Formaldehyde (HCHO)

Molecular Weight 48



Liquid Molar Volume 5881 x 10-2 m3/Kg mol


Gibbs Free Energy -4387.74 KJ/Kg

Heat of Combustion -23077.02 KJ/Kg


Density 1087 Kg/m3

5. Sodium Formate (HCOONa)

Molecular Weight 68



Gibbs Free Energy -8552 KJ/Kg

Liquid Cp 2.247 KJ/Kg K


Solubility Data

Solvent Temp 0C Solubility

g/100gms

solvent

Temp 0C Solubility

g/100gms solvent

Water 25 0C 9.23 97 0C 77.2

Methanol(100%) 25 0C 0.75 50 0C 2.10

Ethanol (100%) 25 0C 0.33 50 0C 1.00

Ether 25 0C 0.45 50 0C 5.00

Ethanol amine 25 0C 0.80 100 0C 10.00

Glycerol 25 0C 16.5 100 0C 44.5

Reactions of Pentaerythritol

1. Oxidation: The polyols are readily attacked by a variety of oxidizing agents like

chromic and nitric acid; it degrades the polyols to CO2.

2. Reduction: The polyols can be reduced by conc. HCL in the presence of certain

catalyst. The end product is secondary alkyl iodide and alkenes without alteration of

the carbon chain.

3. Esterification: Esters with organic acids can be produced in the usual manner with

acid anhydrides or acid chlorides. Esters with inorganic acids are nitrates which are

explosive.

4. Etherification: ether of the polyols are readily produced by reaction with methyl or

ethyl sulfate with the appropriate alkyl chloride and alkali.

5. Nitration: Pentaerythritol on nitration with HNO3 gives Pentaerythritol Tetranitrate

which is an explosive substance.

Grades and Analysis

The analysis of technical pentaerythritol generally includes the following determination:

pentaerythritol contents, melting range, hydroxyl content, ash content, acidity, moisture content,

water solubility, color and physical state. The method used for analysis is the Benzal Method.

Technical pentaerythritol contains 85 to 90 % pentaerythritol and 10 to 15% polyhydric alcohol

such as dipentaerythritol.

Byproduct:

Dipentaerythritol: About 10 to 15 % in the technical grade of pentaerythritol. It is

separated from pentaerythritol by procedure based on differences in solubility in wither

water or alcohol.

Tripentaerythritol: It is separated from pentaerythritol by fractional crystallization.

Sodium formate: It is the main byproduct heaving field about 10%. It is separated from

pentaerythritol by the use of different crystallization temperature. It is used in the

manufacture of formic acid.

Chemistry of Pentaerythritol:

The product pentaerythritol is made from acetaldehyde and formaldehyde in the presence of an

alkaline condensing agent. Initial reaction is three successive molecules of formaldehyde added

to one molecule of acetaldehyde by the classic Aldol reaction.

1. CH3CHO + HCHO HOCH2-CH2CHO

2. HOCH2-CH2CHO + HCHO (HOCH2)2CHCHO

3. (HOCH2)2CHCHO + HCHO HOCH2 | |

HCOH2--------C-------CHO | |

CH2OH

These reactions are truly catalytic and consume no base. They are also reversible. The equlibria of above reactions are displaced continuously to the right by a second possible reaction between two aldehydes in the presence of base i.e. crossed Cannizaro, and the last reaction is not reversible which gives pentaerythritol.

4. CH2OH | | HCOH2--------C-------CHO + HCHO + NaOH (CH2OH) + HCOONa + Na FORMATE | | CH2OH

In the last reaction base, which is a reactant rather than a catalyst is consumed stoichiometrically.

Side Reactions:

Most of the side reactions occur due to formaldehyde and acetaldehyde reacting with each other.

Reactions of acetaldehyde with itself are rapid. However these reactions can be minimized by

maintaining formaldehyde concentration at all times.

Formaldehyde can react with itself to form sugar like products which eventually caramelize to

import color and odor to the products. The self canizzaro of formaldehyde to for CO2 and

methanol can be minimized by avoiding high concentrations of formaldehyde and high

temperature. The desired cannizaro does not progress rapidly below 400C and PH10.

Pentaerythritol is isolated from reaction mixture by successive fractional crystallization.

Product Applications

1. Pentaerythritol is principally used in the surface coating field due to its tetrafunctionality

producing higher viscosity, ability of drying vehicles more rapidly than lower polyols,

hardness, durability, color stability and water resistance. It is usually employed in

combination with polyhydric alcohol such as glycerol to prevent premature gelatin.

2. The resins based on pentaerythritol are used in variety of products such as paints,

varnishes, printing ink and adhesives.

3. Pentaerythritol can be converted to polyether. This compound is characterized by

excellent corrosion resistance and high dimensional stability. It is being used in

manufacturing of molding pipes, gears, valves and metal coating.

4. Pentaerythritol esters of fatty acids are used as plasticizers and also as lubricants.

5. Resins prepared from pentaerythritol such as acrolain is used for electrical insulation,

surface coating, films and fibers.

6. Pentaerythritol is used alone or in combinations with metal salts as heat stabilizers.

7. It is used in the manufacturing of pentaerythritol tetranitrate which is an explosive.

8. It is used in fire retardant surface coatings. When exposed to high temperatures these

coatings froth and swell to produce a solid, non combustible residue which serves to

protect the substrate from fire.

9. Pentaerythritol tetranitrate (PETN) has characteristics similar to those of cyclonite and is

mixed with TNT to form the explosive pentolite. It also forms the core of the explosive

primacord fuses used for detonating demolition charges and the booster charges used on

blasting.

10. Alkyd resin paint is esters of polybasic acids and polyhydric alcohols which have been

modified by oil and fatty acids. Pentaerythritol is used as a polyhydric acid. In particular,

alkyd paint made with pentaerythritol has superior adhersion, weather & water resistance,

color, luster, chemical resistant properties as compared to trihydric alcohols such as

glycerin. They also dry faster than trihydric alcohols.

11. Due to its properties it is also used for many types of varnish enamels.

12. Pentaerythritol and fatty acid esters give plasticity to synthetic resins. Pentaerythritol

tetraacetate, when used as a plasticizer for celluloses, acetates and cellulose esters, gives

them improved plasticity and impact strength.

13. Pentaerythritol and esters of monobasic long chain fatty acids (stearic acid, palmitic

acid), depending on the degree of Esterification, take various physical forms and are used

as emulsifiers in cosmetics, and as abrasives.

14. Monobasic fatty acid esters of pentaerythritol have outstanding characteristics as non-

ionic surface active agents, and are used in softening processes for synthetic fibers and

chemical fibers (acetate, vinylon, nylon). They are also used for smoothing in various

processes of synthetic textile production. Because fatty acid esters of pentaerythritol are

highly stable, have a relatively high boiling point and are low in volatility, they are used

as brake oil as well as lubricating oil and corrosion preventive oil in aircraft engines.

Manufacturers of Pentaerythritol:

Allied Resins and Chemicals Ltd. (India)

Asian Paints (India)

Kanoria Chemicals and Industries Limited. KCIL Group (India)

Perstorp Aegis Chemicals Pvt. Ltd. (India)

Suzhou Fine Chemicals Industrial Group (China)

Shanxi Sanwei Co. Ltd. (China)

AllChem Industries Inc. (USA)

Celanese AG (Germany)

Derivados Forestales SA (Spain)

Polialco SA (Spain)

Dr. Theodor Schuchardt & Company (Germany)

Guizhou Organic Chemical General (China)

Hercules Inc. (USA)

Hubei Yihua Group Limited Liability Company (China)

Mitsubishi Gas Chemical Company Inc. (Japan)

Hunan Zhuzhou Chemical Industry Group Co. Ltd. (China)

JLM Industries Inc. (USA)

Koei Chemical Co. Ltd. (Japan)

Lee Chang Yung Chemical Industry Corporation (Taiwan)

Metafrax JSC (Russia)

Process Selection

Following are the two methods which can be used in the manufacturing of pentaerythritol. These

manufacturing methods are based on the reaction between formaldehyde and acetaldehyde in the

presence of condensing agents.

Method – 1

In this method, pentaerythritol is manufactured by using Soda-ash to remove calcium ion as a

precipitate of CaCO3. Formaldehyde and acetaldehyde are reacted in the presence of Calcium

Hydroxide in S.S. Reactor. The reaction mixture is pumped to precipitator where soda-ash is

added to precipitate CaCO3 leaving formate as sodium formate. After filtration, slurry is

concentrated in a triple effect evaporator. Then crystallization is followed by filtration in the pan

filter. P.E. cake is dissolved and deionised to remove final traces of sodium formate and then the

mixture is again crystallized followed by filtration in second pan filter. The P.E. cake is then

dried and stored.

In this process, recovery of sodium formate and additional P.E. from mother liquor decides the

profit and loss margin. Hence the recovery units are the parts of operation and are not separated

from the main unit in any case. The process requires a large number of equipments made up of

S.S. including two pan filters. This causes a rise in the initial equipment cost, making it

unfeasible for profit making. Another factor which adds to the investment is the requirement of

two separate units for removal calcium carbonate and sodium formate. Considering all these

factors makes this an uneconomical process.

Method – 2

This method uses acids in place soda-ash in the manufacturing of pentaerythritol from

formaldehyde and acetaldehyde. This process is far more economical as compared to the earlier

method. The following main considerations are taken into account for the selection of this

process:

1. The yield in this process is 85 to 90 % which is higher than that in the previous process.

2. Easy removal of sodium ions as they directly combine with formate ions to give sodium

formate.

3. Recent research can help in making the plant operation 99 % automatic.

4. Byproduct sodium formate can be used to produce formic acid which can be reused in the

same plant.

Process Description Method 2

A solution of formaldehyde (20 to 30% by wt.) is added to 50% NaOH solution. While the

temperature is maintained at 15 to 20 0C with suitable agitation, 99% liquid acetaldehyde is

slowly added under the surface of the formaldehyde alkali solution. Since the reaction is

exothermic, external cooling is used to maintain the reaction temperature around 20 to 30 0C.

The mole ratio of formaldehyde to acetaldehyde generally used is between 4.5:1 and 5:1. A ratio

of 1:15 mole hydroxide ion per m ole of acetaldehyde appears to be the optimum amount of

condensing agent.The crude reaction mixture is transferred to the neutralization tank here formic

acid is added to reduce the PH of the solution to 7.8 to 8 and subsequently to remove the sodium

ions present as sodium formate. The solution is then evaporated to a specific gravity of about

1.270. It is then chilled to crystallize pentaerythritol and the resulting slurry is filtered. The

mother liquor goes to the recovery system, where it is reworked and sodium ions are removed as

sodium formate.

The filter cake contains P.E. and Poly-pentaerythritol. The latter material is formed by the side

reaction and is a mixture of either linked polymers such as Di-pentaerythritol ( (CH2OH)3CCH2-

CH2C(CH2OH)3 ) and Tri-pentaerythritol( (CH2OH)3CCH2-O-CH2C(CH2OH)2CH2-O-

CH2C(CH2OH)3 ) and other byproducts found in the reaction. The yield of pentaerythritol is

about 85 to 90% by weight based on the acetaldehyde charged.

Sodium Formate Recovery Unit:

Mother liquor is concentrated in an evaporator crystallizer to the point where sodium formate

comes out of solution. It is separated from P.E. in solution by centrifugation without cooling.

This phase is batch wise with automatically controlled loading and washing cycle. Formate is

dried in smaller but similar equipment to that used in pentaerythritol.

Filtrate from the centrifuge is cooled and P.E crystallizes. After filtering in a press filter, these

crystals are dissolved in water and activated carbon is added. The solution goes through a plate

frame press and then is recycled back to P.E evaporator feed tank.

Thermodynamic FeasibilityCompound ∆HF (KJ/Kmol) ∆G (KJ/Kmol)

CH3CHO -16580.52 128647.64

HCHO -108415.5 -409702.8

NaOH -197610.8 -136567.6

HCOONa -6480332.2 -340609.96

P.E. -764200.42 -553219.656

∆G = ∑ (∆G) Products - ∑ (∆G) Reactants

= (– 553219.656 – 340609.96) – (– 136567.6 – 409702.8 – 128647.64)

= – 218911.576 KJ/Kmol

∆G = –RT ln Ka at T = 298 K

Material Balance

Annual Output: 600 tons

Operation: 3 shifts (8 hours each), 300 working days/year.

Basis: 2000 Kg of P.E. per day

Required Materials:

Formaldehyde (37%) – 6350.00 kg

Acetaldehyde (99%) – 765.00 kg

Sodium Hydroxide (50%) – 2100.00 kg

Formic Acid(34.65%) – 1200.00 kg

Reaction:

4HCHO + CH3CHO + NaOH (CH2OH)4C + HCOONa

Input Data:

Pure HCHO input 6350 X 0.37 = 2350 kg

Pure NaOH input 2100 X 0.5 = 1050 kg

Pure CH3CHO input 765 X 0.99 = 757.5 kg

H2O from HCHO 6350 – 2350 = 4000 kg

H2O from CH3CHO 765 – 757.5 = 7.5 kg

H2O from NaOH 2100 – 1050 = 1050 kg

Hence, Total Water Input 757 + 4000 + 1050 = 5057.5 kg

Output Data:

Moles of CH3CHO = 757.5 / 44 = 17.2 kg.mole.17.2 kg.mole of CH3CHO gives = 17.2 kg.mole of P.E by stoichiometry.Hence P.E. produced = 17.2 X 136.1 = 2341.23 kg

Moles of CH3CHO = 757.5 / 44 = 17.2 kg.mole.17.2 kg.mole of CH3CHO gives = 17.2 kg.mole of HCOONa by stoichiometry.Hence HCOONa produced = 17.21 X 136.1 = 1170 kg

NaOH required = 17.21 X 40 = 688.40 kgNaOH excess = 1050 – 688.4 = 361.6 kgHCHO required = 4 X 17.21 X 30 = 2065.33 kgHCHO excess = 2350 – 2065.33 = 284.67 kg

1. Reactor Material Balance:

Input Output (To Neutralizer)

HCHO 2350 kg

CH3CHO 1050 kg

NaOH 757.5 kg

H2O 5057.5 kg

Total Input 9215 kg

2. Neutralizer Material Balance:

HCOOH + NaOH HCOONa + H2O

Formic acid added = 1200 kg.

P.E. 2341.23 kg

HCOONa 1170 kg

Excess NaOH 361.6 kg

Excess HCHO 284.67 kg

H2O 5057.5 kg

Total Output 9215 kg

Pure formic acid quantity in input = 1200 X 0.346 = 415.84 kg (Since it is 34.6% pure)

Input Neutralizer Output Neutralizer(To holding tank)

P.E. 2341.23 kg P.E 2341.23 kg

HCOONa 1170 kg HCOONa 1783.94 kg

Excess NaOH 361.6 kg H2O 6005.16 kg

Excess HCHO 284.67 kg HCHO 284.67 kg

H2O 5057.5 kg Total Output

10415 kg

HCOOH (34.6%)

1200 kg

Total Input 10415 kg

3. Holding Tank Material Balance:

Holding Tank Input Holding Tank Output (To evaporator)

P.E 2341.23 kg P.E 2341.23 kg

HCOONa 1783.94 kg HCOONa 1783.94 kg

H2O 6005.16 kg H2O 6005.16 kg

HCHO 284.67 kg HCHO 284.67 kg

Total Intput 10415 kg Total Output 10415 kg

4. Evaporator Material Balance:

From Holding Tank To Atmosphere

P.E 2341.23 kg H2O3277.22kg

HCOONa 1783.94 kg HCHO 284.67 kg

H2O 6005.16 kg To Holding Tank II

HCHO 284.67 kg P.E.2341.23 kg

Total Input 10415 kg H2O2727.94 kg

HCOONa1783.94 kg

Total Output 10415 kg

Specific Gravity Volume (Liter) Weight (kg)

P.E 1.396 1677.098 2341.23

HCOONa 1.8 (100%) 991.07 1783.94

H2O 3329 3329

Specific Gravity of Outgoing Slurry = 1.27

1.27 = (wt. of outgoing slurry) / (vol. of outgoing slurry)

= [wt. of (P.E. + HCOONa) + wt. of H2O(x)] / [vol. of (P.E. + HCOONa) + vol. of H2O(x)]

1.27 = [4125.17 + x] / [2668.208 + x]

Hence, x = 2727.94 kg

H2O Evaporated = 6005.94 – 2727.94 = 3277.22kg

5. Holding Tank II Material Balance:

From Evaporator To Vacuum CrystallizerP.E. 2341.23 kg P.E. 2341.23 kg

H2O 2727.94 kg H2O 2727.94 kgHCOONa 1783.94 kg HCOONa 1783.94 kgTotal Input 6853.11 kg Total Output 6853.11 kg

6. Vacuum Crystallizer Material Balance: Yield of P.E. is about 85% to 90%.

Assume that 14.8% of P.E. is lost i.e. will not crystallizer in the crystallizer.

P.E. in solution = 2341.23 X 0.148 = 347.23 kg

Hence, P.E. in crystallizer = 2341.23 – 347.23 = 1994 kg ≈ 2000 kg

From Holding Tank To CentrifugeP.E. 2341.23 kg P.E. Crystal 1994 kg

H2O 2727.94 kg P.E. Liquid 347.23 kg

HCOONa 1783.94 kg H2O 2727.94 kgTotal Input 6853.11 kg HCOONa 1783.94 kg

Total Output 6853.11 kg

7. Centrifuge Material Balance:

From Vacuum Crystallizer To DryerP.E. Crystal 1994 kg P.E. 1994 kg

P.E. Liquid 347.23 kg H2O 127.28 kg

H2O 2727.94 kg To Recovery Unit HCOONa 1783.94 kg HCOONa 1783.94 kgTotal Input 6853.11 kg H2O 2600.66 kg

P.E. 347.23 kgTotal Output 6853.11 kg

Moisture loss with P.E. is assumed 6%

WaterP . E+Water

=0.06=x /(1994+x )

Hence, x = 127.28 kg

8. Dryer Material Balance:

From Centrifuge To BaggingP.E. 1994 kg P.E. 1994 kg

H2O 127.28 kg H2O 6 kgTotal Input 2121.28 kg To Atmosphere

H2O 121.28 kgTotal Output 2121.28 kg

SODIUM FORMATE RECOVERY UNIT:

1. Evaporator Crystallizer Material Balance:

Assume specific gravity of outgoing slurry = 1.37

1.37=wt . of HCOONa+wt . of P . E .+wt .of water (x)

vol . of HCOONa+vol . of P . E .+vol . of water (x)

1.37=1783.94+347.23+x991.07+248.73+x

Hence, x = 1185 kg

Input OutputH2O 2600.66 kg P.E. 347.23 kgP.E. 347.23 kg H2O 1185 kg

HCOONa 1783.94 kg H2O (to atm) 1415.66 kgTotal Input 4731.83 kg HCOONa 1783.94 kg


2. Centrifuge Material Balance:

From Evaporator Crystallizer To CentrifugeH2O 1185 kg H2O. 1113.9 kgP.E. 347.23 kg M.L.P.E. 347.23 kgHCOONa 1783.94 kg HCOONa 89.197 kgTotal Input 3316.17 kg To Atmosphere

HCOONa 1694.743 kgH2O. 71.1 kgTotal Output 3316.17 kg

3. Chiller Material Balance:

From Centrifuge To Filter PressH2O 1113.9 kg H2O 1113.9 kgP.E. 347.23 kg P.E. Crystals 318.86 kgHCOONa 89.197 kg Liquid P.E. 28.37 kgTotal Input 1550.327 kg HCOONa 89.197 kg


4. Filter Press Material Balance:

From Chiller OutputP.E. Crystals 318.86 kg P.E. Crystals 318.86 kgLiquid P.E. 28.37 kg Liquid P.E. 28.37 kgH2O 1113.9 kg H2O 1083.9 kgHCOONa 89.197 kg HCOONa 89.197 kg Total Input 1550.327 kg H2O (to atm) 30 kg


For Recovery Unit:

1. P.E. from dryer having 0.3% moisture = 0.003 X 318.8 = 0.957 ≈ 1

Total P.E from Recovery Unit = 318.8 + 1 = 319.8 ≈ 320 kg

2. Sodium formate from dryer having 0.3% moisture = 0.03 X 1694.7 = 5.08 kg

Total HCOONa = 1694.7 + 5.08 ≈ 1700 kg

3. Total P.E. produced per day = 2000 + 320 = 2320 kg

Energy Balance

1. Energy Balance of Reactor

Reaction: 4CH2O + CH3CHO + NaOH C(CH2OH)4 + HCOONa

Compounds Specific Heat CP (cal/gm0C) Heat of Formation HF (Kcal/gm-mole)

CH2O 0.526 -30.29

CH3CHO 0.54 -41.72

NaOH 0.3125 -112.193

C(CH2OH)4 [P.E.] 0.386 -226.6

HCOONa 0.269 -100

H298 = (HF) products – (HF) reactants = -326 – (-275.11) = -51.49 Kcal/gm-mole

H298 = -51.49 X (2341.23 X 1000) / 136 = -886396.564 Kcal/day

H reactant = ∑ m.Cp.dt = (2350 X 0.526 + 1050 X 0.54 + 757.5 X 0.3125) X (298 – 303)

= -10199.1 Kcal/day

H product = ∑ m.Cp.dt = [(2341.23 X .386) + (1170 X .269)] X 5 = 6092.2 cal/day

Q = H reactant + H product + H298 = -10199.1 + 6092.2 – 886396.564 = -890503.464 kcal/day

Heat loss by radiation 10% = -89050.3464 kcal/day

Heat to be removed = 890903.464 - 89050.3464 = 801453.1176 kcal/day

Q = m.CP.dt = m X 1 X (303-293)

m = 801453.1176 / (10 X 3.78) = 21202.46 gallons

2. Energy Balance of Holding Tank

Temp of Holding Tank = 90 0C

Heat to be supplied = ∑ m.Cp.dt

= (2341.23 X 0.386 + 1783.4 X .269 + 6005.16 X 1 + 284.67 X .576) X (90 – 30)

= 453154.758 kcal/day

Now, considering heat loss due to radiation 10%

Heat required = Heat supplied = 1.1 X 453154.758 = 498470.2338 kcal/day

Heat lost by radiation = 498470.2338 – 453154.758 = 45315.47 kcal/day

Steam Required at 2.1125 atm, Ms = (498470.2338 / 525.6475) = 948.3 kg/day

Heat loss by condensation = 121.27 X 948.3 = 115000 kcal/day

Heat to be supplied by steam = (525.6475 – 121.27) X 948.3

= 383471.19 kcal/day

3. Energy Balance for Dryer

Amount of P.E. to be dried = 1994 kg, Water to be removed = 121.28 kg

Sensible heat required to heat material from 25 0C to 100 0C = ∑ m.Cp.dt

= 1994 X (100 – 75) X .386 + 121.28 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

Hence, Total heat required = 66822.3 + 65491.2 = 132313.5 kcal

Assuming 10% heat loss due to radiation, Hence Heat required = 1.1 X 1323 = 145544.85 kcal

Steam Required = 145544.85 / 525.6475 = 276.88 kg

Enthalpy of Feed = ∑ m.Cp.dt = (1994 X 0.386 + 127.28 X 1) X 5 = 4454.82 kcal

4. Energy Balance of Evaporator

Basis: 2 tons per day

Amount of P.E. present in feed = 2341.23 kg/day

The same amount is to be present in the liquor coming out of the evaporator.

Flow rate:

Total Flow rate (kg/hr) P.E Flow rate (kg/hr) Liquid Flow Rate (kg/hr)

Feed Solution 433.958 97.55 336.40

Thick Liquor 285.54 97.55 167.90

Water + CH2O

evaporated

148.418 168.5

Pressure in atm Temperature in 0F

Inlet stream 2 248

Outlet stream 1 221

Total Temp Drop:

∆T = Th – Tc

Th = 248 0F

Tc = (212 + 192) / 2 = 46 0F

Assume B.P.R. = 1 0F

Effective Temp Drop = 46 – 1 = 45 0F

mf = 10415 kg/day = 433.958 kg/hr

Hf = ∫298

(298+90 )

m. Cp . dt

= (2341.23 X 0.36 + 1783 X .26 + 284.67 X .5760) X (90) = 679708.053 kcal/day

= 28321.17 kg/hr

Hproduct=∫298

398

m .Cp . dt

Hproduct = (2341.23 X 0.38 + 2727.94 X 1 + 1783.94 X 0.269) X 100 = 411152.4 kcal/day

= 17131.39 kcal/hr

V = 3561.89 kg/day = 148.41 kg/hr

Latent heat of water at 1000C = 1150.4 Btu/hr

Hs = 1162.5 Btu/lb = 645.26 Kcal/kg

(Sensible heat + latent heat at 248 0F from Mc Cabe Appendix 6)

Let, mc = X kg/hr

Hc = 188.17 Btu/lb = 105.64 Kcal/kg (sensible heat for condensate at 221 0F)

The Enthalpy Balance Equation is given by: mf.Hf + ms.Hs = (mf – m)Hv + mc.Hc + ms.Hc

28321.17 + X.(645.26) = 17131.35 + 94768.24 + X.(105.64)

Hence, X = 154.8 kg/hr = 3717.20 kg/hr

Steam Economy = (kg water evaporated) / ( kg of steam required) = 3561.89/3717.2 = 0.95

Design of Equipment

Design of Neutralization Tank(ref: Unit operations of chemical engineering: Mc Cabe)

In the neutralization tank formic acid is added to neutralize the excess alkali and to effect

removal of the metallic ion of the condensing agent. Formic cid maybe added to reduce pH of

solution to 7.8 to 8 and subsequently to remove the sodium ion present as sodium formate.

Selection of Equipment: The reaction is carried out in a cylindrical vessel provided with the

drain value at the bottom for the gravity flow. To achieve the good mixing of the reactants, a 6

bladed turbine agitator is provided.

Selection of MOC: In the selection of material of construction for any vessel, the factors to be

considered are initial lost, corrosive action of the reactants, cost of replacement, maintenance and

probable life. Taking the corrosive action of NaOH into account, the best choice to use is

Stainless Steel.

Mechanical Design Calculation: Since the kinetic data is not available, we have to design the

neutralization tank on the basis on reactants mass.

Material in Neutralization Tank Weight in kgs Density at 25 0C (g/cc)

Pentaerythritol 2341.3 1.396

HCOONa 1783.94 1.274

H2O 6005.16 1

HCHO 284.67 1.029

1g/cc = 1 kg/ltr

Average density of reaction mass at 300C = 1.1747 g/cc

Total volume of reaction mass = 10415/1.1747 = 8866.09 ltr/day

Assume residence time in neutralization tank as 2 hours.

Residence time, T = V/ vo

vo = 8866.09 / 24 = 369.42 ltr/hr = 369420 c.c/hr

V = ( ∏ D2 H ) / 4 { D = Diameter of Tank, H = Height of Tank}

Assume H/D = 1.25

Hence, V = [ ∏ D2 (1.25D) ] / 4

D3 = (4V) / ( 3.14 X 1.25) = (4 X 369420) / (3.14 X 1.25) = 376478.98.09 c.c/hr

D = 71.89 cm = 72 cm

H = 1.25D = 1.25 X 72 = 90cm

Shell Thickness: In the tank the pressure is atmospheric, hence the maximum pressure will at the

bottom due to the hydraulic pressure. Taking maximum design pressure to be 20 psi.

Shell Thickness ts = (Pd/2fE) + C

Where, ts = Shell Thickness

f = Allowable stress in psi

E = Joint efficiency

C = Corrosion Allowance

ts = [((20 X 72) / (2 X 20000 X 0.8))+ 0.005] X 2.54

ts = 0.127 cm = 1.27mm

Take shell thickness as 5mm.

We can take thickness of bottom and head equal to shaft i.e. 5mm. Select the elliptical head and

bottom with major to minor axis ratio to be 2. i.e. a = 2b

Inside depth, b = a/2 = I.D/4 = 72/4 = 18cm

Hence, V = ( ∏ / 4 ) X 722 X 90 + ( ∏ X 723 ) / 12 = 464.151 ltr

Space allowance

= [ ( 464.151 – 369.4) / 369.4 ] X 100 = 25 %

Volume of incoming feed – Volume of Head

= ( ∏ / 4 ) X D2 ( h1 – b )

= 369420 – (∏ D3) / 24 = ( ∏ / 4 ) X D2 ( h1 – b )

Hence, h1 = 96.72 cm

This is the height of the liquid level from bottom of elliptical head.

Design of Agitator: (ref: Unit operations of chemical engineering: Mc Cabe)

Calculations are done in order to obtain Power in HP.

In order to provide better mixing, 6 blade turbine agitator is the best choice.

Let diameter of agitator = 0.6 ( Diameter of Tank)

Da = 0.6 X 72 = 43.2 cm

Speed of Agitator = 75rpm

n = 75/60 = 1.25

Assume viscosity of slurry = 5cp

A.V density of slurry = 1.1747 kg/ltr

Nre = (Da X n X ρ) / μ = (43.2)2 X 1.25 X 1.174 / 60 = 54774.144

Nfr = (n2 X Da) / a = 1.252 X 43.2 / 980 = 0.068

From table, a = 1, b = 40

m = ( a - log Nre ) / b = ( 1 – log54774.144) / 40 = -0.0935

P = ((Nfr)m n3Da5) / gc = (0.068)-0.0935 X 1.253 X (43.2/30.48)5 X 1.17 X 62 / 32.17

43.2 cm = 43.2 / 30.8 ft

1.17 kg/ltr = 1.17 X 62 (lb/ft2)

9.81 m/s2 = 32.17 (ft/s2)

Hence, P = 31.95 ft.lbf = 0.05 HP [ 1HP

Taking frictional losses into account, P = 2 HP

Shaft diameter = (53.3 X 2/75)0.33 = 1.123cm [ 2cm

Torque in shaft = (2 X 550) / (2πn) = (2 X 550) / ( 2 X 3.14 X 1.25 ) = 140 ft.lb

Standard Dimensions for Nozzle requirement:

Charging Hole 125 cm diameter

Man Hole 200 X 250 mm diameter

Stuffing Box 100 mm diameter

Drain Value 150 mm diameter

Specification sheet for neutralization tank:

Number Required 1

Type Cylindrical, closed tank with turbine type

agitator

Normal Capacity 370 ltr

Critical Dimension Overall height = 90cm

Inside diameter = 72 cm

Shell thickness = 5cm

Material of construction Stainless steel

Baffles No

Agitator Turbine type

Number of blades = 6

Diameter = 44cm

Rpm = 75

Shaft diameter = 2cm

Material of Construction = Carbon steel

Nozzles Feed charge nozzle = 125 m diameter

Man hole = 200 X 50 mm

Stuffing box stump = 100 mm diameter

Drain value = 150 mm diameter

Supports 4 lug supports, supported at a height of 3m for

gravity flow

Cost Rs 2,00,00 (Approx)

To evaporate 1 lb of water, steam required is about 1 to 1.3 lb or 1000 to 1200 Btu heat is

required.

Evaporator Design Report

Circulation and heat transfer in this type of evaporator is strongly affected by the liquid level as

indicated by an external gauge glass, 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 reduction in 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 room

Easy mechanical descaling

Relatively inexpensive

Mild scaling solution can be used for mechanical cleaning as the tubes are short and large

in diameter

Crystalline products can be removed using the agitator

Procedure followed in design of evaporator assembly: (ref: Unit operations of chemical

engineering: Mc Cabe)

The liquid Flowrate coming out of the evaporator is calculated using material balance. Since the

inlet steam pressure is known, a certain volume is assumed in the evaporator to find the

corresponding temperature for saturated steam. Using this value the total temperature difference

in the evaporator system can be calculated.

The energy balance equations for the evaporator are used in order to calculate the heat duty.

Taking the ud values from Mc Cabe, the total heat transfer area for the evaporator is calculated.

By manipulating the temperature drop appropriate area is calculated. For this calculated area,

tubes requirement for required heat duty is calculated.

Ref: Unit operations of chemical engineering, McCabe.

Q = U.A.∆T

Q = 1316857.143 Kcal/hr

Q = 1316857.143 X 0.252 = 331848 Btu/hr

U = 160 Btu/hr.ft2.0F Ref: Unit operations of chemical engineering, McCabe.

∆T = 45 0F

Hence, A = (331848) / (45 X 160) = 46.09 ft2

For Sheet Diameter: C/S Area of one tube = (π/4).d2.l = (3.14/4) X (2.2)2 X (48) = 182.463 in2

From McCabe, 30% of A is the area of downcomer = 182.463 X 0.3 = 54.4 in2

(π/4).d2.= 54.4 in2

Hence, d ≈ 8 inches

Length & 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 4ft and diameter as

2 inches.

Outer area of each tube = π.do.l = 3.14 X (2.2/12) X 4 = 2.1 ft2

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)/(2 X 0.8 X 3312.5) + 0.08 = 0.2 inches [ minimum

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 across 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 the vapor from the solution. 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 calendria represents the greater part of the

volume taken up by the equipment. The objective is to diminish the risk of entrainment of

droplets of solution projected by boiling. Various MOC used are 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 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 Well (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 the 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 ltr/hr

Working Pressure 2 atm

Critical Dimension Overall height = 220 inches

Inside diameter of Shell = 21.2 inches

Inside diameter of tube = 2 inches

Thickness of shell = 0.2 inches

Thickness of tube = 0.1 inches

Diameter of downcomer = 3 inches

Length of tube = 48 inches

Number of tubes = 22 to 24

Material of construction Stainless steel

Baffles As per requirement

Nozzles or jet stream At steam inlet to increase velocity

Cost Rs 6,00,000(Approx)

List of Major Equipments

1. Reactor

Number Required 1

Type S.S. jacketed cylindrical type reaction vessel

with anchor type agitator.


Capacity of Jacket 890 gallon/hr

Operating conditions P = 1atm, T = 30 0C

Overall dimensions D = 90cm, H = 110cm

Material Of Construction Stainless Steel

Accessories and fittings Thermometer pocket, safety valve,

observation glass assembly, pressure gauge,

manhole etc.

Heat transmitting surface 4 square meter.

Agitator type Anchor type with explosion proof motor

Method of drive: 50 RPM

Electrical Motor Voltage: 415 V

H.P = 7.5, Croft Ratio = 30:1

Cost Rs. 4,00,000 (Approx)

2. Neutralizing Tank

Number Required 1

Type Vertical Cylindrical round bottom tank


Dimensions D = 72cm, H = 90cm, Thickness = 5mm

MOC Stainless Steel

Agitator Turbine type, 6 bladed

Motor speed: 75 RPM, H.P = 3,Voltage = 415 V

Supports Supported at 3m from bottom with 4 legs


3. Centrifuge

Number Required 5

Type Basket Type


Cake Capacity 400 kg/ltr

Overall Dimensions Basket = 60cm I.D , Depth = 30cm,

Outer Case = 61cm I.D.

MOC Basket of S.S plate outer cage lined with S.S

sheet filer of asbestos

Drive Motor H.P = 3, Speed = 1400 RPM, Voltage = 460 V


4. Dryer

Number Required 1

Type Counter current rotary dryer

Normal capacity 150 kg/hr

Dimensions & other details L = 12.5 ft, D = 2.5 ft, Speed = 5 RPM, H.P = 10

Entering Feed Temp = 30 0C

Outgoing Temp = 90 0C

Temp of entering steam = 120 0C

MOC Stainless Steel


5. Evaporator

Number Required 1

Type Short tube calendria type

Normal capacity 1600 ltr

Dimensions & other details D = 55cm, Ht = 550 cm, Shell thickness = 5cm

No of tubes = 24, Ht of tube = 120cm,

D of tube = 50cm, D of Downcomer = 15cm,

thickness of downcomer wall = 5mm, H.P = 10

MOC Stainless Steel


6. Evaporator Crystallizer

Duty To evaporate water from the filtrate and hence

to produce super-saturation with respect to

sodium formate and finally to produce crystals

of HCOONa

No. Required 1

Type Single effect evaporator crystallizer with pump

motor assembly. Heat exchanger vacuum pump

crystallizer with screen.

Capacity 200 ltr/hr

Temperature 100 0C

Pump To handle 200 ltr/hr centrifugal pump


7. Holding Tank

Number Required 2

Type Cylindrical type


Dimensions D = 60cm , H = 75 cm

MOC Stainless Steel

Cost Rs 50,000 (Approx)

8. Pumps

Number Required 8

Type Centrifugal

Normal Capacity 15 Gallons per minute

Drive Motor 3 H.P

MOC Stainless Steel

Cost Rs 50,000 (Approx)

9. Boiler

Duty To produce the steam required in holding tank

to keep the reaction mixture in liquid form.

Number Required 1

Type Vaporax boiler forced circulation coil

Capacity 500 kg/hr

Maximum Pressure 30 p.s.i

Water inlet temperature 30 0C

Firing Liquid Furnace oil

Accessories and mountings Water level indicator, Feed check valve,

Safety valve to control allowable pressure,

Pressure gauge, fusible plug, blow off valve,

Feed pump.

Instrumental process control & precautions

Controllers are gaining utmost importance in industries due to their efficiency, compactness,

response and hazard control capability. These advantages of automatic systems outweigh their

disadvantage of high cost.

Instruments are used for monitoring key process variables during plant operations. Process

variables depend on physical and chemical conditions, which vary with time. To control the

product quality it is necessary that different process variables like temperature and pressure are

maintained to prefix values. Fully automatic plant gives uniform products, avoids accidents and

minimizes the cost of labor; however it causes an increase in the initial cost of the project. the

present project is operated on batch basis and requires moderate capital investment.

Process control and automation together with their instrumentation can be considered as the

mechanical brain and nerves of modern chemical processes. However automatic control is highly

expensive. As far as the current plant is concerned no automation is recommended except for a

temperature controller. The temperature controller is used to control and stop the reaction. Also

the pH in neutralizer is maintained using a controller.

For temperature measurements of the reactor, thermo couples type measuring services are used.

One pressure gauge is also used for the reactor for safety purposes. Various values are provided

in pipe lines for manually controlling the quantity and flow rates of liquids. As the process is

continuous, precise control is needed.

Selection of location & plant layout

One of the most important parts of the final planning is the site location. Careful selection and

engineering research is necessary for attaining the advantage of the process and development

work. Factors contributing towards final site location are as follows:

Raw material supply: Raw material should be cheaply and regularly available. Plant site

near to the raw material source permits considerable reduction in transportation and

storage charges.

Market: Plant location near to the market reduces the cost of product distribution and

shipping time. If the plant is not situated near to the market for its final product, quick

and cheap transportation facilities should be available.

Power and fuel supply: Electricity and fuel should be available regularly otherwise the

production may cease.

Water supply: Process industries require water for processing, cooling, steam generation.

When an industry requires large amounts of water it should be located at areas with

constant water supply.

Climate: Weather has a serious effect on the economic operation of a plant. The

temperature and humidity should be favorable for human body. Pentaerythritol reacts

with oxidizing agents on heating; hence oxidizing atmosphere should be avoided.

Transportation: Waterways, railways and highways are the common means of

transportation. These services should be available near the plant for cheap and quick

transportation.

Labor supply: Availability of labor at stable pay rates should be considered. The other

factors that need to be considered are the intelligence, stability, efficiency of the labor for

economic planning.

Taxes and regulation laws: State and local tax rates on property, income, building codes,

restrictions on transportation are some of the factors which need to considered while

zeroing on the plant location.

Site characteristics: Soil structure, availability of excess space for future expansions and

cost of the site should also be considered.

Layout of the plant: The advantage of gravity flow should be taken if possible in order to

reduce the pumping cost. Water, steam and power should be available for cheap. Safety

of the site location should be considered in order to avoid hazards such as fire explosions.

The site should have a proper waste disposal system. Proper usage of the floor and

elevation space should be planned.

Storage layout: Storage facilities should be provided in isolated or adjoining areas.

Arranging the storage of material so as to facilitate the easy handling is an important

factor which needs to be considered in storage layout.

Equipment layout: While designing the plant layout ample of space should be provided to

all the equipments as they may need replacement, repairs and maintenance at regular time

intervals. If the equipments are placed very close to each other in a plant it raises the risk

of hazards due to interactions of processes. Gravity flow is preferable for viscous

material. In this case reaction mass and also the mass form evaporator.

Service and utilities

The main services required for the manufacturing of Pentaerythritol from formaldehyde and

acetaldehyde are water, steam and electricity.

Process Water: It is required to prepare the solution as per requirements for:

I. Neutralization Reaction

II. Washing the Pentaerythritol cake

III. General services

The requirement can be served by municipal water supply or the industries own water supply is

recommended.

Cooling Water: It is required to extract the exothermic heat of reaction to maintain the

required reaction temperature.

Process Air: The heated air or steam is required in the plant as a heating medium in the

rotary dryer. It is a minor cost item for the process.

Steam: Steam is required in the plant for heating and for evaporation of water in the

evaporator. It is also used in holding tanks to keep the reaction mixture in solution form.

Electricity: When power requirement is high, industries should have their own power

stations. However in small scale industries, power is supplied by the state government

electricity board.

Chilling Water: It is necessary to maintain the temperature at which Pentaerythritol is

crystallized in recovery plant.

Profitability & Cost of Project

A computation of all data relative to the cost of raw materials, land, buildings, labor and

supervision, equipment cost, taxes, insurance, interest etc. should be obtained by the designer on

an approximate pre-construction cost estimation basis. This data can be used for the actual

operation cost accounting if the project goes commercial. It gives the investors a rough estimate

of the total cost of the project and the payback period.

All the costs taken into account may vary according to the market demand and supply.

Basis: 2000kg/day and 300 workings days in a year.

COST ESTIMATION:

(A) Plant Equipment Cost:

Equipment Approximate Cost (Rs. In Lakhs)

Reactor 4

Neutralizer 3

Holding Tank (Quantity: 2) 2 (1 Lakh each)

Evaporator 6

Crystallizer 3

Dryer 8

Centrifuge (Quantity: 5) 15 (3 Lakhs each)

Conveyor 3

Pumps (Quantity: 10) 5 ( 0.5 Lakh each)

Total Cost (A) 49

(B) Recovery Equipment Cost:

Equipment Approximate Cost (Rs. In Lakhs)

Evaporator Crystallizer 6

Chiller 2

Press Filter 3

Centrifuge 3

Total Cost (B) 14

Total Equipment Cost (P.E.C) = Total Cost (A) + Total Cost (B)

= Rs. 49,00,000 +14,00,000 = Rs.63,00,000

Installation Equipment Cost (I.E.C) Rs 6,30,000 (10% of P.E.C)

Piping Cost (P.C) Rs 6,30,000 (10% of P.E.C)

Installation Cost (I.C) Rs 6,30,000 (10% of P.E.C)

Land Cost (2 Acres) Rs 40,00,000 (20000 ft2 X 500 Rs/ft2)

Building Rs. 1,00,00,000

Installation and Electric Fitting (I.E.F) Rs. 9,45,000 (15% of P.E.C)

Total Direct Cost (T.D.C) = I.E.C + P.C + I.C + L.C + Building + I.E.F

T.D.C = Rs. 1,68,35,000

Foundation cost Rs. 5,04,000 (8% of P.E.C)

Platform and support cost Rs. 4,41,000 (7% of P.E.C)

Erection of equipment cost Rs. 9,45,000 ( 15% of P.E.C)

Total cost Rs. 18,90,000

Total Installed Equipment cost (T.E.C) = 18,90,000 + P.E.C + T.D.C = 2,50,25,000

Instrumentation cost Rs. 9,45,000 (15% of P.E.C)

Electrical Insulation cost Rs. 6,30,000 (10% of P.E.C)

Land cost (1 acre) Rs. 20,00,000

Building yard improvement cost & utilities Rs. 2,52,000

Total cost of auxiliary items Rs. 38,27,000

Total physical plant cost (P.P.C) = Total cost of auxiliary items + T.E.C

= Rs. 2,88,52,000

Engineering and construction cost (E.C.C) 1 Rs. 14,42,600 (5% of P.P.C)

Direct Plant Cost (D.P.C) = P.P.C + E.C.C 3,02,98,000

Construction Fee 2 Rs. 6,05,960 (2% of P.P.C)

Contingencies 3 Rs. 30,29,800 (10% of D.P.C)

Total Fixed Investment (F.C.I) = 1+2+3+P.P.C Rs. 3,39,30,360

Working Capital Margin Money 4 Rs. 27,14,428 (8% of F.C.I)

Total Capital Investment (T.C.I) = F.C.I + 4 Rs. 3,66,44,788

Estimation of Product and Raw Material Cost

Formaldehyde = Rs.11/kg X 2350 kg/day X 300 days/yr = Rs. 77,55,000

Acetaldehyde = Rs. 36/kg X 757.7 kg/day X 300 days/yr = Rs. 81,81,000

NaOH (50%) = Rs. 9/kg X 1050 kg/day X 300 days/yr = Rs. 28,35,000

Formic Acid = Rs 9/kg X 415.84 kg/day X 300 days/yr = Rs. 11,22,768

Hence, Total Product and Raw Material Cost is

= 77,55,000 + 81,81,000 + 28,35,000 + 11,22,768 = Rs.1,98,93,768

Labor and Supervision Cost:

General Manager Rs 5 lakhs/annum 1

Engineers Rs. 6 lakhs/annum 2

Chemists Rs. 6 lakhs/annum 3

Clerks Rs. 3 lakhs/annum 3

Skilled Workers Rs. 10.8 lakhs/annum 18 (Rs. 5000/month each)

Semi-skilled Workers Rs. 10.08 lakhs/annum 21 (Rs. 4000/month each)

Unskilled Workers Rs. 10.08 lakhs/annum 30 (Rs. 3000/month each)

Watchmen Rs. 1.44 lakhs/annum 3 (Rs. 4000/month each)

Total Direct Labor Cost Rs. 52.4 lakhs/annum

Utilities:

Cost of Power:

Equipments Power in H.P Quantity

Reactor 7.5 1

Neutralizer 3

Centrifuge 15 (3 H.P each)

Dryer 10

Evaporator 10

Crystallizer 5

Pumps 30 (3 H.P each)

Boilers 10

Lightings, Blowers,

Air compressor, vacuum system

40

Chiller (Recovery equipment) 20

Crystallizer (Recovery equipment) 10

Elevator (Recovery equipment) 3

Filter Press (Recovery equipment) 10

Conveyor (Recovery equipment) 3

Centrifuge (Recovery equipment) 3

Total Power Required 179.6 ≈180

Calculation:

Load 180 KVA

Power Factor 180 KVA X 0.9 162 KVA

Maximum Demand

(80% of operating load)

162 X 0.8 129.6 KVA

Diversity Factor for batch 129.6 KVA X 0.4 51.84 KVA

Total Cost of Power = 51.84 X 0.9 KWHR x 300 days X 24 hrs X 5 Rs/Unit KWHR

= 16.796 lakhs/yr ≈ 16.8 lakhs/yr

Cost of Steam:

Equipment Steam required in kg/day

Evaporator 3800

Dryer 290

Total steam required 4090

Steam required per year = 4090 kg/day X 300 days/yr = 12,27,000 kg/yr

Amount of furnace oil required = 12,27,000 / 13 = 94384.615 kg/yr

Cost of furnace oil to produce steam required

= 94384.615 kg/yr X 16 rs/kg furnace oil = 15.10 lakhs/yr

Cost of water: Water required = 40 m3/day X 300 days/yr = 1200m3/year

Total cost of water = 1200m3/year X 20 rs/m3 = 2.4 lakhs/year

Cost of effluent treatment operation plant = Rs 1.69 lakhs/yr

Utility Cost

Total cost of Power required Rs. 16.8 Lakhs/year

Cost of furnace oil to produce steam required Rs. 15.10 Lakhs/year

Total cost of water Rs. 2.4 Lakhs/year

Cost of effluent treatment operation plant Rs. 1.70 Lakhs/year

Total utility cost Rs. 36 Lakhs/year

Fix charges

Maintenance a Rs. 3 Lakhs/year

Overhead b 10.5 Lakhs/year (20% of D.LC)

Quality control lab & technical assistance c Rs. 5.2 Lakhs/year (10% of D.L.C)

Depreciation cost of 1st year d1 Rs. 40.716 Lakhs/year (12% of F.C.I)

Depreciation cost of each additional year d2 Rs.33.9036 Lakhs/year (10% of F.C.I)

Interest on total investment e Rs. 16.965 Lakhs/year (5% of F.C.I)

Insurance f Rs. 6.786 Lakhs/year (2% of F.C.I)

Total fix charges for 1st year a+b+c+d1+e+f Rs. 83.167 Lakhs/year

Total fix charges for additional year a+b+c+d1+e+f Rs. 76.35 Lakhs/year

Product sales value, Penterythritol = 72 Rs/kg X 2000 kg/day X 300 days/yr= 432 lakhs/year

General expenses, sales, research and development = 2% of 432lakhs = 8.64 lakhs/year

g General expenses, sales, r and d Rs. 8.64 lakhs/year

x Total raw material cost Rs. 198.93768 lakhs/year

y Total direct labor cost (D.L.C) Rs. 52.4 lakhs/year

z Utilities Rs. 36 lakhs/year

F1 Total fix charges for 1st year Rs. 83.167 Lakhs/year

F2 Total fix charges for additional year Rs. 76.35 Lakhs/year

Total manufacturing cost for 1st year (g+x+y+z+F1) 379.14 lakhs

Total manufacturing cost for additional year (g+x+y+z+F2) 372.32 lakhs

Sodium formate recovery (Bypdt) = 15 rs/kg X 1700 kg/day X 300 days/year = Rs. 76.5 lakhs

Total sales of P.E. + sodium formate = 432 + 76.5 = 508.5 lakhs

Sales, Profit & Year

1st year Additional year

x Total sales of P.E. + sodium formate 508.5 lakhs 508.5 lakhs

y Total manufacturing cost 379.14 lakhs 372.32 lakhs

z Gross Profit (x – y) 129.36 lakhs 136.18 lakhs

IT Income tax (35% of Gross Profit) 45.27 lakhs 47.66 lakhs

Net Profit (z – IT) 84.08 lakhs 88.52 lakhs

Payout Period=1+(TC . I−( Net Profit for 1 st Year ) )

Net Profit for additional Year

Payout Period=1+(366.4478−84.08 )

88.52

Payout Period = 4.1898 ≈4.2 years

Conclusion

Pentaerythritol is produced using formaldehyde and acetaldehyde using acids instead of soda-ash

as it is more economical.

Annual Output: 600 tons

Operation: 3 shifts (8 hours each), 300 working days/year.

Basic: 2000 Kg of P.E. per day

Required Materials:

Formaldehyde (37%) – 6350.00 kg

Acetaldehyde (99%) – 765.00 kg

Sodium Hydroxide (50%) – 2100.00 kg

Formic Acid(34.65%) – 1200.00 kg

Based on the production requirements, Material Balance for the entire process and Material

balance for individual equipments is carried out. Sodium Formate is separated from P.E in the

Sodium Formate recovery unit, where it is converted into Formic acid. Energy Balance and

Design of Equipments is studied based on which the requirements and cost of the equipments is

approximated. Cost Analysis helps in determining the pay-back period for the project which

works out to 4.2 years. For the Cost Analysis all the factors which contribute towards the total

expenditure of the project are considered.

References

Mc Ketta J.J., “Encyclopedia Of Chemical Processing and Design. Volume 5”

Perry R.H., “Chemical Engineering Hand Book”

Kirk, Othmer, “Concise Encyclopedia of Chemical Technology”

B.I. Bhatt, S.M. Vora, “Stoichiometry. 4th Edition”

McCabe W.L., Smith J.C., Harriot P., “Unit Operations of Chemical Engineering 5th Edition”

Treybal R.E., “Mass Transfer Operations”

Joshi M.V., “Process Equipment Design”

Peters M.S., Timmerhaus K.D., “Plant Design and Economics for Chemical Engineering 3rd Edition”

Richardson, Coulson “Chemical Engineering Volume 6”

Bhageria Dye Chem Ltd, Vapi, Gujrat

Websites:

www.chemicalland.com

www.chemicalregister.com

www.pentaprocess.com

www.chemicalweekly.com

www.kanoriachem.com

http://www.kanoriachem.com/

http://www.chemicalweekly.com/

http://www.pentaprocess.com/

http://www.chemicalregister.com/

http://www.chemicalland.com/

Manufacturing of Pentaerythritol

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