1

Manufacturing Industrial chemicals

Chandran Udumbasseri

cudumbasseri@yahoo.co.in.

General Introduction

This is a document site where new projects will be published now and then. All the projects

were time tested, implemented and sometimes installed and continued to manufacture. Those

who prefer to adopt can do so.

Projects

1. Industrial oxalic acid that can be manufactured from sugar or such related products.

2. Manufacturing dodecyl benzene sulfonic acid

3. Formulations for detergent powder are also given.

4. Dye intermediate Anthranilic acid from phthalic anhydride

Sugar Oxidation - Oxalic Acid Production

Introduction

Oxalic acid finds use in pharmaceuticals, dye-stuff and textiles industries. Its

derivative diethyl oxalate is the starting material for sulphamethoxazole drug. It is

also used in equipment cleaning, textile finishing, leather tanning and purifying.

Process

Oxalic acid can be produced in pure form (98% +) by the oxidation of cane sugar

products molasses, jaggery and sugar. Mixture of sulfuric acid and nitric acid is

used as oxidizing agent using the catalyst vanadium pentoxide. A 50% sulfuric acid

is used as heat sink and to produce reactive nitryl sulfuric acid which acts as the

intermediate in the oxidation.

The following conditions are used for manufacturing oxalic acid:

1. Conc:Sulfuric acid to Conc Nitric acid ratio = 1.5-2.3 (here for calculation 2.3 ratio is adapted)

2. The glucose to nitric acid mole ratio = 1:3.2 C12H22O11 + H2O 2C6H12O6

2

2C6H12O6 + 12HNO36HOOC-COOH + 12NO + 12H2O

C6H12O6 + 9[O] 3HOOC-COOH + 3H2O

6HNO3 6NO + 9[O] +3H2O

3. Catalyst Vanadium Pentoxide required = 0.05% (m/m) of moles glucose 4. Reaction temperature is maintained at 60-65oC 5. Sucrose is added in 3 hours. 6. Conversion to oxalic acid by this method = 95-98% (~96%) 7. Sucrose is added as 50 % solution in water

Production batch calculation

Target 1200MT per year

Total number of working days per year = 300

Production per day =

= 4000 Kg

Molecular weight of Sucrose = 342.296

Molecular weight of Oxalic acid = 90.036

Molecular weight of Nitric acid = 63.018

Input Quantities

Moles of Oxalic acid, kg mol /day =

= 44.427 kg moles/day

The oxidation reaction shows 6 moles of oxalic acid is produced per mole of

sucrose.

So to produce 44.427 kg moles of oxalic acid the required kg moles of sucrose =

= 7.4 Kg moles of sucrose = 7.4 x 2 = 14.8 Kg moles of glucose.

As the conversion is 96% the required sucrose is =

= 7.713 Kg mole of sucrose.

Weight of sucrose required = 7.713 x 342.296 Kg = 2640 Kg

3

The reaction mole ratio for glucose to Nitric acid is = 1:3.2

For 7.713x2 = 15.426 kg mole of glucose the required kg mole of Nitric acid is

15.426 x 3.2 = 49.36 Kg mole

Weight of Concentrated Nitric acid is = 49.36x 63.018 = 3110.77 Kg

98%Nitric acid =

= 3240 Kg

Conc Sulfuric acid to Conc Nitric acid is 2.3

Weight of concentrated Sulfuric acid is =

= 1352.5 Kg

98% of Sulfuric acid =

= 1380 Kg

Catalyst (Vanadium pentoxide) amount = 0.05% of sucrose moles

For 7.713 Kg mole of sucrose V2O5 required = 0.0039 kg mole = 0.7Kg

(it is sufficient to add catalyst in the first reaction mass if mother liquor is

recycled)

Quantity calculated per day -input

S/N Material description Quantity, Kg

1 Sucrose,50% Solution 5280

2 Nitric acid, 96% 3240

3 Sulfuric acid, 98% 1380

4 Vanadium Catalyst 0.7

Quantity per day -output

Oxalic acid produced = 4000Kg

Nox gas produced

2C6H12O6 + 12HNO36HOOC-COOH + 12NO + 12H2O

1mole of NO at NTP occupies a volume of 22.4 liters.

12 moles has volume 22.4x12 = 269 Liters

4

For 7.713 sucrose the gas volume is 269 x7.713 = 2073 liters

Mother liquor

Nitric acid used = 12 HNO3 = 12x63.018 = 756 kgs per moles of sucrose

For 7.713 mole of sucrose = 756 x7.713 = 1619 =

=1686kg

Un- reacted nitric acid (96%) = 3240-1686 = 1554 Kg

Water from reaction

Water from 12 mole of HNO3 = 12 H2O = 12x18 = 216 Kg

For 7.713 mole sucrose = 216x7.713 = 1666 Kg

Final mother liquor composition

S/N Material description Quantity, Kg

1 Nitric acid 1554

2 Water 1666

3 Sulfuric acid 1380

Total 4600

Nitric acid in mother liquor as calculated = 24 % (in practice only 20% is found)

Specific gravity parameters of all mixtures

Sulfuric acid (98%) 1.84

Nitric acid (96%) 1.49

Sugar solution 1.23

Mother liquor 1.42

Caustic soda (10%) solution (scrubber solution) 1.12

Acid mixture (sulfuric acid/nitric acid) 1.67

Oxalic acid-Bulk density, Kg/m3 833

5

Flow chart

Equipment Design and Selection

Equipments:

1. Reactor vessel 2. Nitric acid addition tank 3. Sugar addition tank 4. Liquid sugar stock tank 5. Sugar pump 6. Air blower 7. Liquid extractor 8. Mother liquor tank 9. Mother liquor pump 10. Nox scrubber 11. Alkaline nitrate tank 12. Dryer 13. Cooling tower 14. Alkali tank 15. Alkali sprayer pump

6

Reactor Volume

Daily cycle = 8hrs (Charging time, ½ hr + Reaction time, 3 ½ hrs + crystallization

time, 3 hrs + unloading reactor content, 1 hr )

Number of cycles per day = 23/8 = 3

Process Parameters

Temperature = 60 – 65oC

Pressure = 1atm

Volume safety factor = 10%

Volume of acid mixture =

= 804 gal/day

(10% excess volume; 3.785Lt = 1gal; 1.67 = sp.gr)

Volume of sugar solution =

= 1248 gal/day

(10% excess volume; 1 gal = 3.785 Lt; sp.gr of sugar solution = 1.23)

Total reaction volume = 804 + 1248 = 2052 gal/day

No of cycles per day = 3

Volume per cycle = 2052/3 = 684 gal/day

Considering the reactor 75% then volume of the reactor needed is =684/0.75 =

912 gallons ~900 gallons

Reactor 900 gallons made of non-corrosive SS

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Nitric acid

Daily requirements of nitric acid =

= 575gal/day

Requirements per cycle = 575/3 = 192gal/cycle

Considering 75% working volume = 192/0.75 = 256 gal/cycle

Nitric acid addition volume ~ 250 gal

Material of tank construction = SS

Sugar addition tank

Daily requirements of sugar solution =

= 1134gal/day

Requirements per cycle = 1134/3 = 378gal/cycle

Considering the reactor 75% then volume of the reactor needed is = 378/0.75 =

504 gal ~500gal

Liquid sugar stock tank

The sugar is prepared for each day once for 3 cycle

Total daily requirement of sugar solution = 1134 gal. Working volume is 75%

Required volume of stock tank = 1134/0.75 = 1512 gal ~1500 gal

Sugar pump

The sugar solution is to be pumped to the addition tank within ½ hr. 378 gal in 30

minutes = 378/30 = 12.6 gpm ~ 10 gpm capacity centrifugal CS pump.

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Nitric acid pump

Volume of nitric acid required per cycle = 192gal/cycle.

The nitric acid is to be pumped within ½ hr, 192/30 = 6.4 gpm ~ 6gpm capacity SS

pump

Air blower

Total amount of NO gas liberated daily = 2073Liters

Gas evolution per minute =

= 1.4 liters/minute

The blower should have capacity to blow 1.5Lt/minutes

Liquid extractor (Liquid solid separator)

The total bulk volume of oxalic acid = 4000Kg

Bulk density of oxalic acid = 833 kg/m3

One cubic meter of oxalic acid weighs 833Kg

4000 kg oxalic acid has bulk volume =

= 4.8 m3

4.8m3 oxalic acid is to be unloaded from the reactor in 1hr. The extractor should

have the capacity to extract 4.8/60 = 0.08 m3/minute = 80Liters/minute

Liquid extractor capacity = 100 liters of solid/minutes

Mother liquor storage tank

Total mother liquor is

9

Volume of sulfuric acid + non reacted Nitric acid + water from sugar solution +

water from oxidation reaction=

+

+

+

= 750 + 1043 + 2640 + 1666 = 6099 Liters

Considering 75% working volume, then

= 8132 liters~8000 Liters

Mother liquor storage tank volume (carbon steel) = 8000 liters =

=2113

gal/day

SS pump is used to pump mother liquor to the reactor.

Nox scrubber (2Nos)

Scrubber of 1 ½ ft diameter, 10ft height, 4-7 gal/min flow with circulating pump;

construction, PP/ Fiberglass.

Alkaline nitrate tank

Nitric oxide has a solubility of 75ml per liter of water. Scrubber can take up to

4gallons per minute. Air blower operates at 2 liter per minute. Nox gas evolution

is 1.5 liters per minute. 100 gallons of 10% caustic alkali is used for scrubbing.

Tank capacity is 150 gallons

Dryer

Bulk volume of oxalic acid produced per day 4.8m3. Hot air circulated tray dryer of

capacity 400Kg (48 trays in each batch of drying) drying per hour at 105oC-110oC.

Total amount per day production is 4000Kg.Total time of drying is 1-1 ½ hrs.

Unloading for 400 Kg is ½ hr. 4000 K of crystals is dried by 20 hrs.

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Cooling tower

The flow of ambient temperature water (25oC) through the jacket of the reactor

should be to maintain a reaction temperature of 60-65oC.

Alkali tank

Tank capacity is 150 gallons. 10% of NaOH solution is used as scrubber solution

Alkaline sprayer pump

Heat of reaction

Reaction

C6H12O6 + 6HNO33HOOC-COOH + 6NO + 6H2O

Heat of formation of glucose = -1268 kJ/mol

Heat of formation of HNO3 = -207.36kJ/mol

Heat of formation of oxalic acid = -822.2 kJ/mol

Heat of formation of NO = +90.25kJ/mol

Heat of formation of water = -241.82 kJ/mol

3HOOC-COOH – (C6H12O6 + 6HNO3- 6NO - 6H2O)

3 ∆Hoxalic acid – (∆Hglucose + 6∆Hnitric acid -6∆HNO - 6∆Hwater)

3(-822.2)-(-1268+6(-207.36)-6(+90.25)-6(-241.82) = -863kJ/mol = -818BTU/mol = -

818000 Btu/kgmol

If this heat is liberated in 3hrs (sugar addition time)

11

=

= 272670

For 3 cycle,

= 90890 BTU/kgmol = 90890x7.713 = 701034 Btu/hr

Assuming a difference of 15oC for cooling water

Mass of water required

m =

=

= 46735 Kg/hr =

= 205 gal/min

Heat transfer area to cool reactor

Inlet water temperature = 25oC

Water outlet temperature = 40oC

Equation to be used, Q = AU∆Tim

Heat transfer coefficient = 45 Btu/ (h) (ft2) (oF)

∆Tlm =

=

= 68oF

Cooling area = A =

=

=229ft2

Equipment specification

1. Reactor vessel, SS316 make = 900gallon 2. Nitric acid addition tank, SS 316 make, 250 gal 3. Sugar addition tank, CS/Al/ PP, 500gal 4. Liquid sugar stock tank, PP/HDPE, 1500gal

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5. Sugar pump, CS, 10gpm 6. Nitric acid, SS, 6gpm 7. Air blower, flow rate, 2 Lt/minute 8. Liquid extractor, SS, 100liters of solid/min 9. Mother liquor tank, SS, 2000gal 10. Mother liquor pump, SS, 11. Nox scrubber, Fiberglass, 1 ½’ x10’ 4 gal/min 12. Alkaline nitrite tank, HDPE, 150 gal 13. Dryer, 48 tray, 500 kg/hr 14. Cooling tower, 15. Alkali tank, 150 gal, HDPE 16. Alkali sprayer pump

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Synthetic Detergents

Introduction

Detergents are those materials that dislodge, remove or disperse solids and liquids from a

surface. All materials that are used for cleaning may be called as detergents. This includes

water, solvents, soap, alkalis and alkaline salts, acids and acidic salts, abrasives, oxidizing and

reducing agents, etc.

Detergents apart from soap are synthetic type detergents. These chemical compounds are

having both hydrophilic and hydrophobic groups. They are characterized and classified into four

groups, anionic, cationic, non-ionic and ampholytic.

Natural detergents are like water, fuller’s earth, neem cake, etc. Synthetic detergents are

chemically prepared or synthesized.

Detergents are surface active materials with characteristics like foaming, emulsifying, reducing

surface tension, etc.

Classification

1. Anionic: they are superior in cleaning action and stable under any conditions 2. Cationic: these types of detergents are weak but they are good in softening, wetting,

foaming, emulsifying. They possess antibacterial properties. 3. Nonionic: they are good detergents but not biodegradable. 4. Amphoteric: they have both acidic and basic properties.

Method of manufacturing

Linear alkyl benzene like dodecyl benzene is sulfonated to give sulfonic acid derivative which on

neutralization gives sodium salt of sulfonic acid.

Additives in detergent formulation

1. Soda ash: neutralizer of acid slurry and keeps the detergent mildly alkaline 2. Tri sodium polyphosphate (TSPP): enhances washing power of detergent and softener

for water 3. Tri sodium phosphate (TSP): increases cleaning power but lower than TSPP 4. Sodium perborate: mild bleaching agent without spoiling color of fabric 5. Carboxy methyl cellulose (CMC): prevents re-deposition of dirt especially for cotton 6. Silicates: corrosion inhibitor and emulsifier

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7. Optical whitening agent: gives brightening appearance to fabric 8. Foam booster: it stabilizes the foam and improves cleaning power of detergent (Saponin

/ lauric ethanol amide) 9. Perfume: gives good odor to detergent 10. Color: copper phthalocyanine gives blue color to the detergent powder

Formulation:

1. Soda ash; 20% 2. cid slurry; 42% 3. STPP; 36% 4. CMC; 2% 5. Blue dye; 0.0001% 6. Whitening agent 7. Perfume

Processing

1. Take acid slurry in the blender 2. Add soda ash slowly under mixing (allow carbon dioxide to escape) 3. Add STPP and CMC under mixing 4. When mixing is complete, add brightener and color mixing thoroughly to homogenize 5. Dry in air circulating oven and pulverize to get homogenized powder. Spray drying gives

satisfactory product. 6. Too much powdered product is not acceptable to customer.

Quality of detergent

The commercial detergent has a pH value of 9.6. Higher pH is unsuitable for silk and woolen

fabrics. Higher pH is acceptable for terylene, nylon, and cotton.

Other types of formulations

A

1.1. Acid slurry; 18% 1.2. Soda ash;22% 1.3. Borax; 25% 1.4. Tri sodium phosphate; 22% 1.5. CMC; 2% 1.6. STPP; 11%

B

2.1. Dodecyl benzene sodium sulfonate (DBSS); 30%

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2.2. STPP; 30% 2.3. Sodium silicate; 10% 2.4. Sodium perborate; 5% 2.5. Sodium tolyl sulfonate; 3% 2.6. Benzotriazole 2.7. Sodium sulfate and water as filler

C

3.1 Dodecyl benzene sodium sulfonate; 20% 3.2. STPP; 30% 3.3. Sodium silicate; 10% 3.4. Sodium perborate; 8% 3.5. CMC; 1% 3.6. Sodium tolyl sulfonate; 2% 3.7. Alkanolamide; 3% 3.8. Brightener 3.9. Sodium sulfate and water as filler

Kitchen scoring powder

4.1. Magnesium silicate; 90% 4.2. DBSS; 4% 4.3. STPP; 2% 4.4. Sodium sulfate and water as filler

Commercial detergent powder

5.1. DBSS; 30%

5.2. STPP; 25%

5.3. Sodium silicate; 0.25%

5.4. Brightener; 0.25%

5.5. Perfume; 0.1%

5.6. CMC; 2%

5.7. Builders and fillers 42.4%

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Detergent bar

6.1. DBSS; 34%

6.2. Nonyl phenol ethoxylate; 4%

6.3. Sodium silicate (43.5%); 10%

6.4. Magnesium silicate; 2%

6.5. Polyvinyl alcohol+ polyvinyl acetate; 0.8% (coating solution)

Enzymatic detergent bar

7.1. DBSS; 33% 7.2. Soda ash; 27% 7.3. Corn Starch; 30% 7.4. Water; 8% 7.5. Sodium bisulfite; 0.6% 7.6. Protease; 0.6% 7.7. Color; 0.3 7.8. Perfume; 0.2

Liquid Detergent

7.9. Acid slurry; 4.5% 7.10. Potassium polyphosphate; 5% 7.11. Potassium silicate; 2% 7.12. Potassium xylene sulfonate; 2% 7.13. Diethanolamine; 1.7% 7.14. CMC; 0.5% 7.15. Brightener; 0.01% 7.16. Water to make up

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Project Evaluation-Sodium Dodecyl benzene sulfonate (Detergent)

Process: Benzene is alkylated with dodecene using AlCl3 catalyst by Friedl-Crafts alkylation

method. The temperature is maintained at 115oF. Dodecene is separated from un-reacted

benzene and dodecene by distillation. Dodecylbenzene is sulfonated with Oleum (20%).

Dodecylbenzene sulfonic acid is neutralized with 20% NaOH solution. Na2SO4 is the by product.

Overall conversion is 85% while alkylation is taken as 90% and sulfonation as 95%.

1. Sulfonation is carried out by 20% Oleum and the ration of alkylbenzne to Oleum is 1.25. 2. Spent sulfuric acid is removed by adding 0.244Kg to every 1.25 Kg of 20% Oleum. 3. 20% NaOH solution with 25% is used for neutralizing the spent acid. 4. Reaction duration is 6 hrs

Flow Chart

Production target 12000 MT (12,000,000 Kg)

Alkylation yield 85-95%

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Sulfonation 100%

Neutralization 95%

Ratio of Oleum to Dodecyl benzene 1.25

Spent acid removal

For each 0.244 lb of water require 1.25 lb of 20% Oleum

A 25% excess of 20% NaOH for neutralization

Molecular weights

Benzene 78.1

Dodecene 168.3

Dodecyl benzene 246.4

Dodecyl benzene sulfonic acid 325.4

Sodium Dodecyl benzene sulfonate 348.5

Material Balance calculation

Calendar days of production 300

Product 85% (DBS acid + H2SO4 (excess) + NaOH -> Na DB sulfonate

+ Na2SO4)

Daily production

= 97.56 Kg mol/day

Considering both yields 95% (sulfonation) and 90% (alkylation)

The requirement of benzene =

= 114.1 Kg mol/day

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Quantity of benzene = (114.1)*(78.1) = 8911 Kg/day

Requirement of Dodecene = 114.1 Kg mol/day = (114.1)*(168.3) = 19203 Kg/day

Total = 8911 +19203 = 28114 Kg/day

20% Oleum requirement = (1.25)*(28114) = 35143 Kg/day

Water requirement =

= 6860 Kg/day

Caustic soda (20%) for neutralization = (1.25) (35143) = 43929 Kg/day

Catalyst (AlCl3) = (0.05) (28114) = 1405 Kg/day

Requirements Table per day

S/N Item Qty, lb/day

1 Benzene 8911

2 Dodecene 19203

3 20% Oleum 35143

4 Water 6860

5 Caustic soda (20%) 43929

6 Catalyst, AlCl3 1405

Alkylation

Alkylate yield (0.9)(114.1)(246.4) = 25303 Kg/day (Mol wt of Dodecyl benzene =

246.4)

20

Unreacted Benzene = (0.1)(8911) = 891. Kg/day

Unreacted dodecane = (0.1)(19203) = 1920 Kg/day

Sulfur balance

After neutralization 85% is Sodium Dodecyl Benzene sulfonate and 15% sodium sulfate

Sulfur in =

= 12017 Kg/day

Sulfur out =sulfur in detergent + sulfur in spent acid

Sulfur out in detergent =

+

= 3132 + 1356 = 4488 Kg/day

(Daily production of detergent = 12000000/300 = 40000 Kg)

Sulfur out in spent acid = 12017 – 4488 = 7529 Kg/day

Weight of 78% H2SO4 = (7529)

= 29498 Kg/day

Material Balance:

Benzene +Dodecane+H2SO4 +NaOH +H2O = 8911 + 19203 + 35143 + 43929 + 6860 = 114046

Dodecyl Benzene + Unreacted Benzene + Unreacted Dodecane

25302 + 891 + 1920 ……. (Recovered un-reacted material 891+ 1920 =

2811)

Dodecyl Benzene + H2SO4

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25302 + 35143 = 60445

+ Water

60445 + 6860 = 67305

Spent acid (78%) = 29498

Separated sulfonate layer = 67305-29498 = 37807

+ Caustic (20%) added

37807 + 43929 = 81736

Water removed = 41736 kg

Detergent = 40000 Kg

Equipment Design and Selection

Reactor Volume

Daily cycle = 6 hrs

No of cycle per day 24/6 = 4

Process parameters

Temperature = 115oF

Pressure = 1 atm

Volume safety factor = 10%

Volume of benzene used per day =

= 2942 gal/day

(10% excess, 3.785 Lt = 1Gal; 0.88 density of benzene in g/cc)

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Volume of Dodecene used per day =

= 7408 gal/day

(10% excess, 3.785Lt = 1gal; 0.7533 density of dodecene in g/cc)

Volume of AlCl3 used per day =

= 152 gal/day

(Density of 2.44 g/cc, 62.4 convert to lb/cu ft; 7.48 cu ft to lb/gal conversion)

Total volume = volume of (Benzene + Dodecene + AlCl3) = 2842.86+ 7408.46+ 152 = 10503

gal/day

Volume per cycle = 10503/4 = 2625.89gal/cycle

If the reactor is 75% full on each cycle, the volume of the reactor needed is 2625.89/0.75 =

3501 gal

Reactor volume can be 3500 gallon

Heat of Reaction:

C6H6(l) + C12H26 (l) -> C6H5.C12H25(l)

Heat of reaction is equal to heat of formation of dodecyl benzene - heat of formation of

benzene –heat of formation of dodecane

∆Hr = ∆Hf(C6H5.C12H25)l - ∆Hf(C6H6)l - ∆Hf(C12H24)l

∆Hf(C6H5.C12H25)l = -54348cal/g mol

∆Hf(C12H24)l = -51239 cal/g mol

∆Hf(C6H6)l =11717 cal/g mol

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∆Hr = -54348-11717-(-51239) = -14826 cal/g mol = -58840 Btu/Kg mol

If this heat is liberated in 3 hrs (of 4 hr cycle)

Q = -58840

)

= -279728.5 Btu/hr

Assume temperature difference 10oF for cooling water

Mass of water required to remove heat of reaction,

m =

=

= 27972.8 Kg/hr =

= 123 gal per minute (gpm)

So the volume flow is 123 gpm approximatelty 125 gpm

The transfer pump should have flow capacity 125 gpm (carbon steel)

Heat transfer area to cool reactor

Assume water inlet 80oF

Let the rise in temperature is 10oF

Equation to be used is, Q = A U ∆Tlm

The overall heat transfer coefficient (HTC) (U) for this type of heat transfer may be taken as 45

Btu/ (h) (ft2) (oF)

[Usually the HTC is measured in W/(m2)K. For conversion 1 W/(m2)K = 0.1761 Btu/ (h) (ft2) (oF)]

45/0.1761 = 255 W/(m2)K

Average temperature difference can be calculated using the following equation

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∆Tlm =

= 29.7oF

U = 45 Btu/ (h) (ft2) (oF)

So cooling area, A =

=

=209ft2

Total reactor volume is 3500 gal =3500x0.13368 cu ft = 468 cu ft

Storage area

Benzene = 2843 x 4 = 11372 gal

Dodecyl benzene = 7408 x 4 = 29632 gal

So select 12,000 gal CS tank for benzene and 30000 gal CS tank for dodecyl benzene

Pump size

Calculating for 10 minute filling

Benzene pumping =

= 71 gpm (~75 gpm)

Dodecyl benzene =

= 185 gpm (~190 gpm)

Alkylate transfer from holding tank to fractionators =

= 7 gpm. This pumping is

continuous to the fractionator.

Select 10 gpm displacement pump

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Equipment specification

1. Benzene storage tank, CS = 12000 gal 2. Dodecyl benzene storage tank, CS=30000 gal 3. Holding tank for intermediate alkylates, CS = 11000 gal 4. Benzene transfer (centrifugal) pump, CS = 75 gpm 5. Dodecyl benzene transfer (centrifugal) pump, CS = 190 gpm (or two pumps of 95gpm) 6. Cooling water (centrifugal) pumping to jacket, CS = 125 gpm (or two pumps of 65gpm) 7. Alkylate transfer (displacement) pump, CS = 10 gpm 8. Reactor volume (alkylation) = 3500 gal

26

Dye Intermediate – Anthranilic Acid

Introduction

Anthranilic acid finds use in dyes, pigments, perfumes and medicine. It is the starting

material for the sedative methaqualone and diuretics furosemide. Its esters are used as

perfumes and corrosion inhibitors.

Process

Phthalic anhydride is reacted with aqueous ammonia at 230oC to give phthalimide.

Phthalimide is then oxidized with sodium hypochlorite to give anthranilic acid

+ NH3

O

NH

O

O

NH2

O

ONa

NaOH Hydrolysis

Cl2, NaOH

O

NH

O

ONa

Cl-Cl

N

O

ONa

O

+H2O N

O

ONa

O

H

OH

- CO2NH2

O

O

OH

NH2

O

OH

+O

ONaOCl {Cl

2 , [O]}

Acidification

NaOH Hydrolysis

NaOCl {Cl2 , [O]}

Acidification

NH2

O

OH

O

NH

O

O

O

O

O

NH2

O

ONa

N

O

ONa

O

O

NH2

O

ONa

O

NH

O

27

Method

1. Phthalic anhydride is charged in to the Amidation reactor which is heated externally by thermic fluid. Add aqueous ammonia (28% or 16%) to the reactor. Heat the content to 300oC. The liberated ammonia should be scrubbed after condensing the liberated gas.

2. Water is allowed to evaporate out along with ammonia gas. The mass becomes a hot melt (Melting at 232-235oC) with the separation of water as condensate. The hot melt is phthalimide. Discharge the product and cool to room temperature and crush the mass to fine powder.

3. Prepare 18% sodium hydroxide solution, cool 10oC and charge to oxidation reactor. Add the Phthalimide powder and dissolve keeping the temperature at 10-15oC. The addition of phthalimide should be carried out at a fast rate (less time for phthlimide hydrolysis).

4. Cool the mixture to 10oC and start adding sodium hypo chlorite (10% available chlorine) solution at a fast rate taking care not to increase the temperature above 15oC.

5. Allow the temperature to increase to 70-80oC once the oxidation is complete. As the temperature go up CO2 evolves from the mixture.

6. When the gas evolution ceases, check the pH and neutralize the mixture to crystallize out anthranilic acid. Filter off the crystals and dry

Production Batch Calculation

Target 1000 Kg/day

Conversion:

Phthalic anhydride to phthalimide = 96%

Phthalimide to anthranilic acid = 88%

Molecular weight of Anthranilic acid = 137.136

No of Moles per day =

= 7.292 moles/day

Molecular weight of Phthalic anhydride = 148.12

Theoretical qty = 7.292 x 148.12 = 1080.09

Considering conversion,

Required amount of Phthalic anhydride =

= 1279 kg/day

Taking purity of Phthalic anhydride as 97%

28

Feed qty =

= 1318Kg /day

Molecular weight of ammonia = 17.034

For every mole of Phthalic anhydride 2.4 mole of ammonia is practically used

For

= 8.635 moles

For 8.635 moles required ammonia is 8.635x2.4 = 20.72 moles

Required 28% aqueous ammonia solution =

= 1260 Kg/day

For hydrolysis to 2-amido benzoic acid sodium salt

For every mole of Phthalimide 2.4 mole of NaOH is used

For 8.635 moles, NaOH required = 8.635 x 2.4 = 20.72 moles

Required 30% aqueous NaOH solution =

= 2762 Kg/day

For every mole of sodium salt 1mole of molecular chlorine is required for oxidation

Required sodium hypo chlorite with 10% chlorine =

= 7345 Kg/day

NaOH in NaOCl solution: usually the NaOCl of specific gravity 1.2 is having a minimum

pH value of 11.86. This corresponds to free NaOH of 0.025% by weight

7345 Kg of NaOCl solution contains 1.84 Kg of NaOH = 0.046Kg moles

Total NaOH = NaOH added + NaOH from NaOCl = 20.72moles + 0.046 = 20.766 Kg

moles

20.766 NaOH requires =

= 3785 Kg of 20% HCl solution for neutralization

Reaction time

1. Amidation reaction Raw material charging time =1hr Reaction time = 4hrs Cooling to RT and discharging = 2 hrs Total = 7hrs

29

2. Hydrolysis and oxidation Charging to reactor = 1 hr Oxidation = 2 hr Discharging and neutralization = 2 hr Filtration of crystals of anthranilic acid = 1hr

Quantity of Raw Material – input /per day

S/N Material Description Qty, Kgs

1 Phthalic anhydride (purity=97%) 1318

2 Ammonia (28%) solution 1260

3 NaOH (30%) solution 2762

4 NaOCl (10%) solution 7345

5 HCl (20%) for neutralization 3785

Specific gravity

Ammonia solution (28%) = 0.9

NaOH (30%) = 1.328

NaOCl (10%) = 1.173

HCl (20%) = 1.01

Volume of the dilute solutions- input per day

S/N Material Description Qty, Lts

1 Ammonia solution (28%) 1400

2 NaOH solution (30%) 2080

3 NaOCl solution (10%) 6262

4 HCl solution (20%) 3748

30

Flow Chart:

Equipment Design and Selection

1. Amidation Reactor with distillation/condenser 2. Oxidation Reactor 3. Crystallizer 4. Neutralizer 5. Heater 6. Chiller

Equipment Volume

1. Amidation Reactor with distillation/condenser

Feed volumes:

Phthalic anhydride (specific gravity = 1.53) = 1318 Kg =

= 861 Lts

Ammonia (28%) solution = 1400 lts

Total feed volume = 861 + 1400 = 2261 Lts

Working volume 75%, then required volume =

= 3015 ~ 3000Lts

Expected Phthalimide = 7.292 x 147.12 = 1072 Kg

If the conversion is 96%, then 1072x0.96 = 1030 Kg

Construction = MS

2. Oxidation Reactor

Feed volume

31

Feed from amidation reactor = 1072Kg

NaOH (30%) solution = 2080 Lts

NaOCl (10%) solution = 6262 Lt

Total feed volume = 2260 + 2080 + 6262 = 10602 Lts

Working volume 75%, then required volume =

= 14136 Lts~ 14200Lts

Construction = MS

3. Crystallizer

3000Lts

Construction = MS

4. Neutralizer

14200 Lts

Construction = HDPE

5. Heater

Mass of material = 1318 Kg + 1260 Kg = 2578 Kg

Temperature difference 25oC to 235

oC

Heat capacity of 2.69kJ/Kg

6. Chiller

Heat of reaction

Heat of formation of Phthalic anhydride = -460.37 kJ/mole

Heat of formation of phthalimide = -317.9kJ/mole

Heat of formation of anthranilic acid = -401.08 kJ/mole

Heat of formation of ammonia (aq) = -80.8 kJ/mole

Heat of reaction = Σ ∆Hof (products) - Σ ∆H

of (reactants)

Reaction 1

O

O

O

+ NH3

O

NH

O

Heat of reaction= (-317.9 kJ/mole) - (-460.37kJ/mole + -80.8 kJ/mole) = -317.9 +

541.17 = +223.27 kJ/mole

= +223270 J/mole =

Btu/Kg mole = + 211.63 Btu/Kg mole

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Reaction is endothermic. So heating is required.

The reaction mass has to be heated up to 230oC from RT.

Total moles = 7.292 mole

Total heat = =211.63x7.292 = 1543 Btu in 4 hrs = 385 Btu/hr

Q = m Cp ∆T

∆T = 230-25 = 205oC

m = 1318 + 1260 = 2578 Kg

Reaction 2

O

NH

O

NaOH Hydrolysis

NaOCl {Cl2 , [O]}

Acidification

NH2

O

OH

Heat of reaction = Σ ∆Hof (products) - Σ ∆Ho

f (reactants)

= (-401.08)-(-317.9) = -83.18 kJ/mole = -83180 J/mole =

= -78.84 Btu/mole

Equipment specification

1. Amidation Reactor with condenser, MS, 3000Lts

2. Oxidation Reactor, MS, 14200Lts

3. Crystallizer, MS, 3000Lts

4. Neutralizer, HDPE, 14200Lts

5. Chiller

6. Heater

7. Sodium hypochlorite plant