New Calculation
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Transcript of New Calculation
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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
COMPONENTFLOWRATE (kg/day)COMPONENTFLOWRATE (kg/day)
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