Manufacturing of Pentaerythritol

MANUFACTURING OF PENTAERYTHRITOL

By Siddharth Maity B.E., Mumbai University, 2006

An Industrial Engineering Project Report Submitted to the Faculty of the Speed Scientific School of the University of Louisville in Partial Fulfillment of the Requirements for the Degree of

Master of Engineering in Industrial Engineering

Department of Industrial Engineering University of Louisville Louisville, Kentucky December 2009

MANUFACTURING OF PENTAERYTHRITOL By Siddharth Maity B.E., Mumbai University, 2006

An Industrial Engineering Project Report Approved 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

References

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 properties 1. Pentaerythritol

Structure:

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

Physical form

White, crystalline and odourless

Molecular weight

136.15

Composition

C = 42.2%

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

H = 8.8%

O = 49%

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

Boiling Point

293.59 K

Melting Point

150.19 K

Liquid Molar Volume

5.652 x 10-2 m3/Kg mol

Heat of Formation

-3769.3 KJ/Kg

Gibbs Free energy

-2923.81 KJ/Kg

Heat of Fusion

73.07 KJ/Kg

Heat of Combustion

-25054.02 KJ/Kg

Liquid Cp

2.483 KJ/Kg-K

Density

1015.7 Kg/m3

4. Formaldehyde (HCHO)

Molecular Weight

48

Boiling Point

235.49 K

Melting Point

129.9 K

Liquid Molar Volume

5881 x 10-2 m3/Kg mol

Heat of Formation

-5471.53 KJ/Kg

Gibbs Free Energy

-4387.74 KJ/Kg

Heat of Combustion

-23077.02 KJ/Kg

Liquid Cp

2.343 KJ/Kg-K

Density

1087 Kg/m3

5. Sodium Formate (HCOONa) Molecular Weight

68

Melting Point

533.04 K

Heat of Formation

-9529.9 KJ/Kg

Gibbs Free Energy

-8552 KJ/Kg

Liquid Cp

2.247 KJ/Kg K

Density

1918.9 Kg/m3

Solubility Data Solvent

Temp 0C

Solubility

Temp 0C

g/100gms

Solubility g/100gms solvent

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. CH2OH | | HCOH2--------C-------CHO + HCHO + NaOH | | CH2OH 4.

(CH2OH) + HCOONa + Na FORMATE

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 CO 2 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 nonionic 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)3CCH2CH2C(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 Feasibility Compound

∆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 kg NaOH excess = 1050 – 688.4 = 361.6 kg HCHO required = 4 X 17.21 X 30 = 2065.33 kg HCHO 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

Pure formic acid quantity in input = 1200 X H2O Total Output

0.346 = 415.84 kg (Since it is 34.6% pure) Input Neutralizer

5057.5 kg 9215 kg

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 3277.22k g 284.67 kg

P.E HCOONa

2341.23 kg 1783.94 kg

H2O HCHO

H2O

6005.16 kg

HCHO

284.67 kg

Total Input

10415 kg

To Holding Tank II 2341.23 P.E. kg 2727.94 H2O kg 1783.94 HCOONa 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

3329

3329

H2O

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 P.E.

2341.23 kg

To Vacuum Crystallizer P.E. 2341.23 kg

H2O HCOONa Total Input

2727.94 kg 1783.94 kg 6853.11 kg

H2O HCOONa Total Output

2727.94 kg 1783.94 kg 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 P.E. 2341.23 kg

To Centrifuge P.E. Crystal

1994 kg

H2O

2727.94 kg

P.E. Liquid

347.23 kg

HCOONa Total Input

1783.94 kg 6853.11 kg

H2O HCOONa Total Output

2727.94 kg 1783.94 kg 6853.11 kg

7. Centrifuge Material Balance: From Vacuum Crystallizer P.E. Crystal 1994 kg

To Dryer P.E.

1994 kg

P.E. Liquid

347.23 kg

H2O

127.28 kg

H2O HCOONa Total Input

2727.94 kg 1783.94 kg 6853.11 kg

To Recovery Unit HCOONa H2O P.E. Total Output

1783.94 kg 2600.66 kg 347.23 kg 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 P.E. 1994 kg

To Bagging P.E.

H2O Total Input

H2O To Atmosphere H2O Total Output

127.28 kg 2121.28 kg

1994 kg 6 kg 121.28 kg 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 H2O P.E.

2600.66 kg 347.23 kg

Output P.E. H2O

347.23 kg 1185 kg

HCOONa Total Input

1783.94 kg 4731.83 kg

H2O (to atm) HCOONa Total Output

1415.66 kg 1783.94 kg 4731.83 kg

2. Centrifuge Material Balance: From Evaporator Crystallizer H2O 1185 kg P.E. 347.23 kg HCOONa 1783.94 kg Total Input 3316.17 kg

To Centrifuge H2O. M.L.P.E. HCOONa To Atmosphere HCOONa H2O. Total Output

1113.9 kg 347.23 kg 89.197 kg 1694.743 kg 71.1 kg 3316.17 kg

3. Chiller Material Balance: From Centrifuge H2O P.E. HCOONa Total Input

1113.9 kg 347.23 kg 89.197 kg 1550.327 kg

To Filter Press H2O P.E. Crystals Liquid P.E. HCOONa Total Output

1113.9 kg 318.86 kg 28.37 kg 89.197 kg 1550.327 kg

4. Filter Press Material Balance: From Chiller P.E. Crystals Liquid P.E. H2O HCOONa Total Input

318.86 kg 28.37 kg 1113.9 kg 89.197 kg 1550.327 kg

Output P.E. Crystals Liquid P.E. H2O HCOONa H2O (to atm) Total Output

318.86 kg 28.37 kg 1083.9 kg 89.197 kg 30 kg 1550.327 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 148.418

168.5

evaporated

Inlet stream

Pressure in atm

Temperature in 0F

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= 298398m.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.

Normal Capacity

850 ltr

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

Normal Capacity

370 ltr

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

Cost

Rs 2,00,000 (Approx)

3. Centrifuge

Number Required

5

Type

Basket Type

Normal Capacity

85 ltr

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

Cost

Rs 5,00,000 (Approx)

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

Cost

Rs 8,00,000 (Approx)

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

Cost

Rs 6,00,000 (Approx)

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

Cost

Rs 1,00,000 (Approx)

7. Holding Tank Number Required

2

Type

Cylindrical type

Normal Capacity

160 ltr

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

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

(15% of P.E.C)

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

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

Rs. 27,14,428

4

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

(5% of P.P.C)

(8% of F.C.I)

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,

40

Air compressor, vacuum system 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

162 X 0.8

129.6 KVA

129.6 KVA X 0.4

51.84 KVA

(80% of operating load) Diversity Factor for batch

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 Depreciation cost of 1st year

c

d1

Depreciation cost of each additional year

Rs. 5.2 Lakhs/year (10% of D.L.C) Rs. 40.716 Lakhs/year (12% of F.C.I)

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 1st YearNet Profit for additional Year Payout Period=1+366.4478-84.0888.52

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