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WaterHydrolysis process

Acid treated + Palm oilPalm oil

H2SO4HydrolysisR-100

Figure 1: Process flow of hydrolysis

Table 1: Summary of mass balance for hydrolysis processINPUTOUTPUT

COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)

Palm oil

Water(S-104)

H2SO43200

1000

480

(S-101)Acid treated+Palm oil

4680

total46804680

ConclusionHydrolysis is the first process in production of acetone-butanol-ethanol(ABE).The raw material is palm oil which is then hyrdrolysed with H2SO4 as the catalyzed to produce acid treated and palm oil. This reaction is non-reactive.

NaOHNeutralization Process

Glucose + Na2SO4

Neutralization R-101Acid treated + palm oil

Figure 2 : Process flow of Neutralization

Table 2 : Summary of mass balance for eutralization processINPUTOUTPUT


Acid treated+Palm oil4680Glucose+Na2SO4

5071.8

NaOH 391.8

Total5071.8

5071.8

Conclusion The product from output of hydrolysis process will undergo neutralization. In this process, NaOH was add into ( acid treated + palm oil ) to form (glucose + Na2SO4).). this reaction is non-reactant.

Water(S-122)Filtration Process

FiltrationF-100Glucose Glucose + Na2SO4 (S-116)

Solid (S-126)

Figure 3 : Process flow of Filtration

Table 3 : Summary of mass balance for filtration processINPUTOUTPUT


Glucose+Na2SO4

5071.8Glucose( S-127)

4201.032

Water(S-122) 400Solid (S-126)

1270.77

Total5472.8

5471.802

Conclusion The product of neutralization will undergo filtration process. During this process the filtration of Na2SO4 is occur.the additional of water is needed to conduct filtration process.as result the glucose( S-127) and solid(S-126) form.this reaction is reactive.

Sterilization Process

Glucose (S-117)

Sterilization E-101Glucose (S-127)

Figure 4 : Process flow of SterilizationTable 4: Summary of mass balance for sterilization processINPUTOUTPUT


Glucose(S-127)4201.032Glucose(S-117)

4201.032

Total4201.032

4201.032

ConclusionThis is process flow of sterilization. In this process only glucose(S-127)is undergo to sterilization process. The product of this process is glucose(S-117). This reaction is non-reactive.

Fermentation Process

Ammonium(S-125)

Glucose FermentationOutlet(S-117) (R-102)(S-102)Water

Inoculum(S-132)Figure 5: Process Flow of Fermentation

Table 5: Summary of mass balance of fermentation processINPUTOUTPUT


Glucose(S-117)Ammonium(S-125)Inoculum(S-132)Water 4201.032

200

320

628Outlet(S-102)5178.12

Total5349.0325178.12

ConclusionThe product from sterilization process will then will undergo fermentation process. In this process, we will glucose,ammonium and inoculum will be fermented and will produce the outlet.This is a reactive reaction.

MANUAL CALCULATION FOR ENTHALPHY OF FERMENTATION PROCESS

InputOutput

Component Mass flow rate(kg/day)Enthalphy Component Mass flow rate(kg/day)Enthalphy

Glucose 4201Ammonia1878

Ammonia 200Ammonium acetate19960

Water 628Water 615

Inoculums320Carbon dioxide689.12

Open System

where

Cp of

Inlet

Outlet

Filtration Process

Water (S-106)

(S-124)

Filtration F-101Inlet (S-102)

Biomass (S-115)

Figure 6 : Process flow of Filtration

Table 6 : Summary of mass balance for filtration processINPUTOUTPUT


Inlet (S-102)5178.12(S-124)

4281.0424

Water(S-106)4000Biomass(S-115)4897.6988

Total9178.12

9178.12

ConclusionAfter fermentation process the product will undergo filtration process. The inlet(S-102) will added with water. Then it will produce (S-124) and Biomass(S-115). This is non-reactive reactant.Storage 1

S-124 Storage TankS-103V-106

Figure 7: Storage tank

Table 7: Summary of mass balance of storage tank.INPUTOUTPUT


S-1244281.0424S-1034281.0424

total4281.04244281.0424

ConclusionThe product from filtration process will be stored in storage tank.There will be no reaction in this process thats why the input is the same with output. Therefore, this reaction is non-reactive.

Stripping Process

Nitrogen(S-108)

S-103 StrippingS-107 (D-100)

Waste(S-111))Figure 8:Process Flow of Stripping ProcessTable 8: Summary of mass balance of stripping processINPUTOUTPUT


S-103

Nitrogen(S-108)4281.0424

30.63

S-107

Waste(S-111)102.64

4209.01

total4311.674311.65

ConclusionThe product from storage tank will then undergo stripping process.This process is where the product from storage will be added with nitrogen.Then, it will produce some waste and new product.This reaction is a reactive reaction.

Condensation Process

N2 (S-112)

S-107

CondenserS-110E-103

Cold waterFigure 9: Process flow of condensation process

Table 9: Summary of mass balance of condensation processINPUTOUTPUT


S-107102.64S-110N2(S-112)72.0330.61

total102.64102.64

ConclusionThe product from stripping process will then undergo condensation. It will release N2 as a waste.Since the output value is differ from input value, thus we can conclude that this reaction is a reactive reaction.

1St Distillation Process

Acetone

1st Distillation(D-101)S-110S-121

Figure 10: Process flow of 1st Distillation ProcessTable 10: Summary of mass balance of 1st Distillation ProcessINPUTOUTPUT


S-11072.03

S-121

Acetone44.35

27.68

total72.0372.03

ConclusionThe product from condensation process will be undergo first distillation that will produce some product along with Acetone. Since this is a non-reactive process, therefore the input and output does not change.

2nd Distillation Process

Ethanol

2nd Distillation(D-102)S-119S-121

Figure 11: Process Flow of 2nd DistillationTable 11: Summary of mass balance of 2nd distillation processINPUTOUTPUT


S-12144.35

Ethanol

S-11917.45

26.9

total44.3544.35

ConclusionThe product from first distillation will undergo second distillation which is then produced Ethanol and other product.This reaction is not reactive.

Manual calculation for distillation process:

Molar fraction (input stream) :i) S-121 = 44.35 kgmolecular weight of = x g/molassume x=46 g/mol n = (44.35 kg) /(46 g/mol) x (1000 g/kg)=964.13 mol

Molar fraction (output stream) :Upper layer:i) ethanol = 17.45 kgmolecular weight of = 86.17 g/moln = (17.45 kg)/(86.17 g/mol)x(1000g/kg)= 202.51 mol

ii) S-119 = 26.9 kgmolecular weight of S-119 = x g/molassume x=54 n = (26.9 kg)/(54g/mol)x(1000g/kg)= 498.148mol

Total molar fraction, xfxf= output / input = (202.51+498.148) mol / (964.13) mol = 0.726 mol

3rd Distillation Process

Butyric Acid

3rd Distillation(D-103)S-119S-118

Figure 12: Process Flow of 3rd Distillation ProcessTable 12: Summary of mass balance of 3rd distillation processINPUTOUTPUT


S-119

26.9

Butyric Acid

S-1187.288

19.612

total26.926.9

ConclusionThe other product apart of Ethanol from second distillation will then undergo third distillation process to produce butyric acid and other product. Again this is a non-reactive reaction.

Mixer Process

Mixer(M-100)S-123Butanol + Acetic Acid

Figure 13: Process Flow of MixerTable 13: Summary of mass balance of Mixing processINPUTOUTPUT


Butanol+Acetic Acid12.864

6.748S-12319.612

total19.61219.612

ConclusionIn this process, Butanol will be mixed with acetic acid to form a product.This is a non-reactive reaction.

Decanting Process

Decanting(D-104)S-123Butanol

Acetic Acid

Figure 14: Process Flow of Decanting ProcessTable 14: Summary of mass balance of decanting processINPUTOUTPUT


S-12319.612Butanol

Acetic Acid 12.63

6.9748

total19.61219.612

ConclusionLastly, the product from mixing process will be entering decanting process. In this process, Butanol is produced along with Acetic Acid. This is a non-reactive reaction.

Xkg C4H10O/kg(1-X)kg C2H4O2/kg19.6120.7kg C4H10O/kg0.3kg C2H4O2/kg6.7480.4kg C4H10O/kg0.6kg C2H402/kg12.864Mass Balance calculation in mixing process

Mixer

BUTANOL BALANCE(12.864x O.4kg C4H10O/kg) + (6.748x 0.7kg C4H10O/kg) = (19.612x Xkg C4H10O/kg)X = 0.503

ACETIC ACID BALANCEInput = output(12.864x 0.6kg C2H4O2/kg) + (6.748x 0.3kg C2H4O2/kg) = (19.612x (1 0.503)9.7428kg C2H4O2/kg = 9.7428kg C2H4O2/kg

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