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ECONOMICAL RECOVERY OF ETHYL ACETATE AND TETRAHYDROFURAN
FROM THE PRODUCTION OF CEFUROXIME ACID
V.RAGHUNATH 11707203026N.SESHA GOPALAN 11707203038
INTERNAL GUIDE : S.JENISH EXTERNAL GUIDE : SATISH
ABOUT ORCHID CHEMICALS & PHARMACEUTICALS LTD.• Orchid Chemicals & Pharmaceuticals Ltd (Orchid) was established
in 1992 as a 100% Export Oriented Unit (EOU). Commencing operations in 1994, Orchid has achieved amazing and consistent growth, quantitatively and qualitatively to emerge among the Top-15 companies in the Indian pharmaceutical industry in a short span of fifteen years of operations.
• Orchid has an established end-to-end connected infrastructure for drug discovery and development. Orchid has two manufacturing sites for APIs (at Alathur near Chennai and at Aurangabad, near Mumbai) and three manufacturing sites for Dosage forms (at Irungattukottai and Alathur in Chennai), besides a world-class R&D centre (at Sholinganallur, Chennai).
• In Alathur, Separation of complex solvent mixtures, residual from production processes, are done in order to get single high purity fractions to be reused or to be sold with economical benefit.
Multipurpose solvent recovery plantMixture Pre-treatment plant
ABSTRACT
• Recovery of complex solvent mixtures in the production of pharmaceutical components is very complicated.
• This is due to the presence of TDS and solvents mixtures having the tendency to form azeotropes.
• Ethyl Acetate and TetraHydroFuran, are the by-products during the production of cefuroxime acid.
• The solvents are recycled & recovered in order to get single high purity fractions to be reused or to be sold with economical benefit.
• Several operations like distillation, flashing, dehydration are involved here.
• The percentage recovery & percentage loss for every operation involved is studied and estimated.
ETHYL ACETATE
• Ethyl acetate is the ester of ethanol and acetic acid; it is manufactured on a large scale for use as a solvent.
• CAS No: 141-78-6 • Properties Molecular formula : C4H8O2 • Molar mass : 88.105 g/mol • Appearance : colorless liquid • Density : 0.897 g/cm³, • Melting point : −83.6 °C, 190 K, -118 °F• Boiling point : 77.1 °C, 350 K, 171 °F• Solubility in water : 8.3 g/100 mL (20 °C)• Refractive index (nD) : 1.3720 • Viscosity : 0.426 cP at 25 °C• Vapour density: 3.04• Vapour pressure: 100 mm Hg at 27 C • Flash point: -4 C
TETRAHYDROFURAN
• IUPAC name : Oxolane• CAS No.: 109-99-9 • Molecular formula : C4H8O • Molar mass : 72.11 g/mol • Appearance : colorless liquid • Density : 0.8892 g/cm3 (g/mL) @ 20 °C,• Melting point : −108.4 °C (164.75 K)• Boiling point : 66 °C (339.15 K)• Solubility in water Miscible Viscosity : 0.48 cP at 25 °C• Vapor Density (Air=1): 2.5 • Vapor Pressure (mm Hg): 129 @ 20C (68F) • Evaporation Rate (BuAc=1): 8.0 • Flash point : −14 °C
RECOVERY PROCESS
REMOVAL OF TDS EA AND THF ENRICHMENT RECOVERY OF EA RECOVERY OF THF
PROCESS DESCRIPTION
• REMOVAL OF TDS FLASHING • EA AND THF ENRICHMENT DISTILLATION • RECOVERY OF EA
DEHYDRATION USING CaCl2 DISTILLATION (THF REMOVAL)• RECOVERY OF THF DISTILLATION (EA REMOVAL) DEHYDRATION USING NaOH DISTILLATION (ANHYDROUS THF RECOVERY)
NOMENCLATURE
EA = Ethyl Acetate THF = Tetrahydrofuran OI = Other Impurities MC = Moisture Content F = Feed R = Residue T = Top Product B = Bottom Product V = Vent CaCl2 = Calcium Chloride CAS = Chemical Abstract Service
MASS BALANCE
STUDY Calculation of amount of Feed, Top product, Bottom
Product/Residue in every operation. Analysis of Composition of the above, in terms of Percentage. The major components to be separated are Ethyl Acetete &
TetraHydroFuran and other components are termed as Other Impurities.
The composition is analysed through chromatography, The peak value is noted.
The amount of EA, THF, OI & MC in Feed, Top product, Bottom Product/Residue in every operation is calculated.
Loss in flashing/dehydration is mainly by residue. Loss in distillation is mainly attributed to loss by vent owing
to the fact that the column is closed & continuous. The percentage recovery of THF & EA for every process is
calculated The percentage loss of THF & EA for every process is
calculated
TDS REMOVAL
FLASHING
TOP TEMP.=80 OC
BOTTOM TEMP.=100 OC
TOP PRODUCT = 7150 LEA = 52 %THF = 43.5 %OI = 4.5 %MC = 4 %MOTHER LIQUOR = 7530 L
EA = 51 %THF = 43 %OI = 6 %MC = 4 %
RESIDUE = 380 L
MASS BALANCEF = T + RFxf = Txt + Rxr
% of EA in Residue :7530*(51/100) = 7150*(52/100) + 380*(x/100)3840.3 = 3718 + 3.8xx = 32.18%
% of THF in residue :7530*(43/100) = 7150*(43.5/100) + 380*(x/100)3237.9 = 3110.25 + 3.8xx = 33.59%
% of OI in residue :7530*(6/100) = 7150*(4.5/100) + 380*(x/100)451.80 = 321.75 + 3.8xx = 34.22%
RESULTEA THF OI MC
MOTHER LIQUOR ( FEED)
3840.30 3237.9 451.80 301.20
TOP PRODUCT 3718 3110.25 321.75 286
RESIDUE 122.3 127.65 130.35 15.2
Percentage of recovery of EA = ((3718/3840.3)*100) = 96.8%Percentage of recovery of THF = ((3110.25/3237.9)*100) = 96.05%
Percentage of loss of EA in residue = ((122.3/3840.3)*100) = 3.18%Percentage of loss of THF in residue = ((127.65/3237.9)*100) = 3.94%
EA AND THF ENRICHMENT
CONTINOUS COLUMN
TOP TEMP.=64.5 OCBOTTOM
TEMP.=70OC
TOP PRODUCT = 2281.25 LEA = 3.5 %THF = 92.65 %OI = 3.85 %MC = 4 %
BOTTOM PRODUCT = 4372.91 LEA = 78 %THF = 19.2 %OI = 2.8 %MC = 3.6 %
FEED = 7150 LEA = 52 %THF = 43.5 %OI = 4.5%MC = 4 %
MASS BALANCEF = T + B + VFxf = Txt + Bxb + Vxv
% of EA lost by vent7150*(52/100) = 2281.5*(3.5/100)+4372.91*(78/100) +
495.86*(x/100)x = 45.889%
% of THF lost by vent7150*(43.5/100) = 2281.5*(92.65/100)+4372.91*(19.2/100) +
495.86*(x/100)x = 31.622%
% of OI lost by vent7150*(4.5/100) = 2281.5*(3.85/100)+4372.91*(2.8/100) +
495.86*(x/100)x = 22.484%
RESULTEA THF OI MC
FEED 3718 3110.25 321.75 286
TOP PRODUCT 79.84 2113.57 87.82 91.25
BOTTOM PRODUCT 3410.61 839.85 122.44 157.42
VENT 227.55 156.83 111.49 37.33
Percentage of recovery of EA = ((3490.45/3718)*100) = 93.83%Percentage of recovery of THF = ((2953.16/3110.25)*100) = 94.94%
Percentage of loss of EA by vent = ((227.55/3718)*100) = 6.12%Percentage of loss of THF by vent = ((157.09/3110.25)*100) = 5.05%
EA ENRICHMENT
RECOVERY OF ETHYL ACETATE DEHYDRATION USING CACL2
DEHYDRATION USING CaCl2
PRODUCT = 8950 LEA = 80 %THF = 15.2 %OI = 4.8 %MC = 1.35 %
RESIDUE = 549.35 L + 415.65 L (MC)
FEED = 9500 L + 532 L (MC)CACL2 = 390 LEA = 78.5 %THF = 1.15 %OI = 6.8%MC = 5.6 %
MASS BALANCEF = T + RFxf = Txt + Rxr
% of EA in Residue :9500*(78.2/100) = 8950*(80/100) + 549.35*(x/100)7429 = 7160 + 5.49xx = 48.99%
% of THF in residue :9500*(1.15/100) = 8950*(15.2/100) + 549.35*(x/100)1425 = 1360.4 + 5.49xx = 11.76%
% of OI in residue :9500*(6.8/100) = 8950*(4.8/100) + 549.35*(x/100)646 = 429.6 + 5.49xx = 39.415%
EA THF OI MC
FEED 7429 1425 646 532
PRODUCT 7160 1360.4 429.6 116.35
RESIDUE 269 64.6 216.4 415.65
RESULT
Percentage of recovery of EA = ((7160/7429)*100) = 96.37%
Percentage of loss of EA in residue = ((269/7429)*100) = 3.62%
RECOVERY OF ETHYL ACETATE DISTILLATION (THF REMOVAL)
DISTILLATION COLUMN
TOP TEMP.=64.5 OC
BOTTOM TEMP.=70.4 OC
TOP PRODUCT = 1691 L + 40.58 (MC) LEA = 7.1 %THF = 78.5 %OI = 14.4 %MC = 2.4 %
BOTTOM PRODUCT = 6809 L + 7.48 (MC) LEA = 99.46 %THF = 0.20 %OI = 0.33 %MC = 0.11 %
FEED = 8950 L + 116.35 (MC) LEA = 80 %THF = 15.2 %OI = 4.8%MC = 1.35 %
MASS BALANCEF = T + B + VFxf = Txt + Bxb + Vxv
% of EA lost by vent8950*(80/100) = 1691*(7.1/100)+6809*(99.46/100) + 450.71*(x/100)x = 59.39%
% of THF lost by vent8950*(15.2/100) = 1691*(78.5/100)+6809*(19.2/100) + 450.71
*(x/100)x = 4.29%
% of OI lost by vent8950*(4.8/100) = 1691*(14.4/100)+6809*(0.33/100) + 450.71*(x/100)x = 36.30%
RESULT
EA THF OI MC
FEED 7160 1360.4 429.6 116.35
TOP PRODUCT 120.06 1327.43 243.5 40.5
BOTTOM PRODUCT 6772.23 13.61 22.46 7.48
VENT 267.71 19.36 163.34 68.37
Percentage of recovery of EA = ((6892.29/7160)*100) = 96.26%
Percentage of loss of EA = ((267.71/7160)*100) = 3.73%
THF ENRICHMENT
RECOVERY OF TETRAHYDROFURAN DISTILLATION - (EA REMOVAL)
DISTILLATION COLUMN
TOP TEMP.=64OC
BOTTOM TEMP.=70OC
BOTTOM PRODUCT = 4030 LEA = 8.17 %THF = 89.80 %OI = 2.03 %MC = 3.5 %
TOP PRODUCT = 12660 LEA = 0.02 %THF = 99.28 %OI = 0.7 %MC = 5 %
FEED = 17593 LEA = 2.13 %THF = 96.76 %OI = 1.11 %MC = 6.3 %
F = T + B + VFxf = Txt + Bxb + Vxv
% of EA lost by vent17593*(2.13/100) = 12600*(0.02/100)+4030*(8.17/100) +
902.03*(x/100)x = 59.39%
% of THF lost by vent17593*(96.76/100) = 12600*(99.28/100)+4030*(89.80/100) +
902.03*(x/100)x = 92.48%
% of OI lost by vent17593*(1.11/100) = 12600*(0.7/100)+4030*(2.03/100) +
902.03*(x/100)x = 2.75%
MASS BALANCE
RESULT
EA THF OI MC
FEED 374.73 17022 195.28 1108.359
TOP PRODUCT 2.53 12568.84 88.62 633
BOTTOM PRODUCT 329.25 3618.94 81.80 141.05
VENT 42.95 834.22 24.86 334.309
Percentage of recovery of THF= ((16187.78/17022)*100) = 95.09%
Percentage of loss of THF = ((834.22/17022)*100) = 4.9%
RECOVERY OF TETRAHYDROFURAN DEHYDRATION USING CAUSTICS
DEHYDRATION USING
CAUSTICS FLAKES
RESIDUE = 500.61 L
FEED = 12660 LEA = 0.035 %THF = 99.28 %OI = 0.69 %MC = 4 %
TOP PRODUCT = 12160 LEA = 0.02 %THF = 99.33 %OI = 0.45 %MC = 0.25 %
F = T + RFxf = Txt + Rxr
% of EA in Residue :12660*(0.035/100) = 12160*(0.02/100) + 500.61*(x/100)4.43 = 2.43 + 5.00xx = 0.4%
% of THF in residue :12660*(0.035/100) = 12160*(0.02/100) + 500.61*(x/100) 12568.8 = 12078.5 + 5.00xx = 98.06%
% of OI in residue :12660*(0.035/100) = 12160*(0.02/100) + 500.61*(x/100) 87.35 = 79.04 + 5.00xx = 1.662%
MASS BALANCE
RESULTEA THF OI MC
FEED 374.73 17022 195.28 1108.359
TOP PRODUCT 2.53 12568.84 88.62 633
BOTTOM PRODUCT 329.25 3618.94 81.80 141.05
Percentage of recovery of THF= ((12078.5/12660)*100) = 96.09%
Percentage of loss of THF = ((490.3/12660)*100) = 3.9%
RECOVERY OF TETRAHYDROFURAN DISTILLATION (THF RECOVERY)
TOP PRODUCT = 575.42 LEA = 0.1 %THF = 96.2 %OI = 0.7 %MC = 3.4%
BOTTOM PRODUCT = 7200 L (ANHYDROUS THF)EA = 0.017 %THF = 99.74 %OI = 0.243 %MC = 0.04 %
DISTILLATION COLUMN
TOP TEMP.=63OC
BOTTOM TEMP.=68OC
FEED = 8165.4 LEA = 0.02 %THF = 99.33 %OI = 0.65 %MC = 0.5 %
F = T + B + VFxf = Txt + Bxb + Vxv
% of EA lost by vent8165.1*(0.02/100) = 575.42*(0.1/100)+7200*(0.017/100) +
407.73*(x/100)x = 0.076%
% of THF lost by vent8165.1*(99.3/100) = 575.42*(96.2/100)+7200*(99.74/100) +
407.73*(x/100)x = 92.18%
% of OI lost by vent8165.1*(0.65/100) = 575.42*(0.7/100)+7200*(0.243/100) +
407.73*(x/100)x = 7.7389%
MASS BALANCE
RESULT
EA THF OI MC
FEED 1.63 8110.69 53.07 40.82
BOTTOM PRODUCT 1.224 7181.28 17.496 19.56
TOP PRODUCT 0.09 553.55 4.02 2.88
Percentage of recovery of THF= ((7734.83/8110.69)*100) = 95.36%
Percentage of loss of THF = ((375.86/8110.69)*100) = 4.6%
ENERGY BALANCE
STUDY Calculation of amount of Feed, Top product, Bottom
Product/Residue in every operation. Analysis of Composition of the above, in terms of
Percentage. The energy supplied initially calculated. The energy out with the product is also calculated. The difference between the energy in and energy
out is the energy lost. Energy loss is due to 3 reasons Barebody losses,
Insulation losses, Natural condensation.
TDS REMOVAL
TOP TEMP = 83.2 C
BOTTOM TEMP =
84.5
84.5 - 35
83.2 - 55
83.2 - 50
6969.76 Kg
EA - 51% = 3417.86THF - 43% = 2881.73MC - 4% = 301.20
EA - 52% = 2987.4THF - 43.5% = 2499.1MC - 4% = 239.38
EA - 52% = 341.3THF - 43.5% = 285.51MC - 4% = 38.59
ENERGY BALANCEmλ+mCpΔT
ENERGY IN:
Qf = (6294.59*0.5*49.5) + (3417.86*87.6) + (2881.73*98) + (268.068*19.5) + (268.08*540)
Qf = 895755.014 KCal
ENERGY OUT:
Qc1 = (5486.5*0.5*28.2) + (2987.4*87.6) + (2499.1*98) + (239.38*28) + (239.38*540)
Qc1 = 719983.406 KCal
Qc2 = (626.81*0.5*33.2) + (341.30*87.6) + (285.51*98) + (38.54*33.2) + (38.54*540)
Qc2 = 90402.64 KCal
Qc1 +Qc2 = 810386.1 KCal
RESULTQf – (Qc1 + Qc2) = 85368.91 Kcal
The net loss in flashing stage is 85368.91 Kcal
EA – THF ENRICHMENT
TOP TEMP.=64.5 OC
CONTINOUS COLUMN
BOTTOM TEMP.=70OC
77.8 - 35
78 -
55
79 -
77
55 -
35
64 - 55
64 - 50
ENERGY BALANCEmλ+mCpΔT
ENERGY IN:
Qf = (1152*0.5*42.8) + (599.04*87.6) + (501.12*98) + (48*1*42.8) + (48*540)
Qf = 154170.6 KCal
Qr = (4369.92*0.5*7.2) + (152.32*87.6) + (4048.7*98) + (182.08*7.2) + (182.08*540)
Qr = 525449.32 Kcal
Qi = Qf+ Qr = 679619.38 KCal
ENERGY BALANCE
ENERGY OUT:
Qc1 = (4320*0.5*9) + (151.5*87.6) + (4002.4*98) + (180*9) + (180*540)
Qc1 = 523740.3 KCal
Qc2 = (480*0.5*14) + (16.8*87.6) + (444.72*98) + (120*14) + (120*540)
Qc2 = 59494.24 KCal
Qc3 = (430.08*0.5*20) + (17.92*20)
Qc3 = 8960 KCal
Qphe = (966*0.5*23) + (36.072*23)
Qphe = 11938.6 KCal
Qo = Qc1 +Qc2 +Qc3 +Qphe = 604133.196 KCal
RESULTQi – Qo = 75486.184 Kcal
The net loss in EA & THF enrichment is 75486.184 Kcal
EA ENRICHMENT
THF REMOVAL
TOP TEMP.=68 OC
CONTINOUS COLUMN
BOTTOM TEMP.=80OC
78 - 35
80 -
55
80 -
72
55 -
35
68 - 55
68 - 50
ENERGY BALANCEmλ+mCpΔT
ENERGY IN:
Qf = (1183*0.5*43) + (947.4*87.6) + (179.93*98) + (16.2*43) + (116.2*540)
Qf = 135521.68 KCal
Qr = (4694.56*0.5*8) + (333.31*87.6) + (3685.22*98) + (115.44*8) + (115.44*540)
Qr = 472388.876 Kcal
Qi = Qf+ Qr = 607910.566 KCal
ENERGY BALANCE
ENERGY OUT:
Qc1 = (4392*0.5*13) + (311.8*87.6) + (3447.92*98) + (1108*1*13) + (108*540)
Qc1 = 453462.24 KCal
Qc2 = (448*0.5*18) + (34.64*87.6) + (351.68*98) + (12*18) + (12*540)
Qc2 = 48227.10 KCal
Qc3 = (185.44*0.5*20) + (4.56*20)
Qc3 = 1945.6 KCal
Qphe = (2746.98*0.5*25) + (3.02*25)
Qphe = 34412.75 KCal
Qo = Qc1 +Qc2 +Qc3 +Qphe = 538047.69 KCal
RESULTQi – Qo = 69862.56 Kcal
The net loss in THF removal is 69862.56 Kcal
THF ENRICHMENT
EA REMOVAL
TOP TEMP.=63.5 OC
CONTINOUS COLUMN
BOTTOM TEMP.=69.2OC
71 - 35
69.2 - 55
71 -
63
55 -
35
63.8 - 55
68.3 - 50
ENERGY BALANCEmλ+mCpΔT
ENERGY IN:
Qf = (1124.4*0.5*36) + (23.94*87.6) + (1087.96*98) + (75.6*43) + (75.6*540)
Qf = 172426.42 KCal
Qr = (3800*0.5*8) + (76*87.6) + (3772.64*98) + (200*8) + (200*540)
Qr = 494585.2 Kcal
Qi = Qf+ Qr = 667011.716 KCal
ENERGY BALANCE
ENERGY OUT:
Qc1 = (4275*0.5*8.8) + (0.85*87.6) + (4244.22*98) + (225*8.8) + (225*540)
Qc1 = 558078.02 KCal
Qc2 = (475*0.5*13.8) + (0.09*87.6) + (471.58*98) + (25*13.8) + (25*540)
Qc2 = 63345.224 KCal
Qc3 = (950*0.5*20) + (50*20)
Qc3 = 10500 KCal
Qphe = (579*0.5*14.2) + (21*14.2)
Qphe = 4409.1 KCal
Qo = Qc1 +Qc2 +Qc3 +Qphe = 636332.34 KCal
RESULTQi – Qo = 30679.370 Kcal
The net loss in THF removal is 30679.370Kcal
ANHYDROUS THF RECOVERY
TOP TEMP.=65.3 OC
CONTINOUS COLUMN
BOTTOM TEMP.=70OC
66 - 35
70 - 55
78 -
63
55 -
35
65.33 - 55
65.33 - 50
ENERGY BALANCEmλ+mCpΔT
ENERGY IN:
Qf = (1194*0.5*31) + (0.23*87.6) + (1186*98) + (6*31) + (6*540)
Qf = 138181.1 KCal
Qr = (3960*0.5*8) + (3.96*87.6) + (3812.07*98) + (139.4*8) + (139.4*540)
Qr = 466163.356 Kcal
Qi = Qf+ Qr = 604344.456 KCal
ENERGY BALANCE
ENERGY OUT:
Qc1 = (4347*0.5*10.3) + (4181.81*98) + (4.37*87.6) + (153*10.3) + (153*540)
Qc1 = 516780.5 KCal
Qc2 = (483*0.5*15.3) + (464.64*98) + (1.48*87.6) + (17*540) + (17*15.3)
Qc2 = 58711.8 KCal
Qc3 = (72.45*0.5*20) + (2.55*20)
Qc3 = 775.5 KCal
Qphe = (1199.28*0.5*15) + (0.72*15)
Qphe = 9005.4 KCal
Qo = Qc1 +Qc2 +Qc3 +Qphe = 585273.2 KCal
RESULTQi – Qo = 19071.2 Kcal
The net loss in THF removal is 19071.2 KCal
POWER CONSUMPTION
STUDY Power consumption in the recovery process is
mainly due to the usage of agitator in the flashing and dehydration process.
The power consumed is determined by finding the capacity of the motor used for agitation.
From the motor capacity, the power consumed for the specific batch is found out.
REMOVAL OF TDS Motor capacity = 22 KW Motor error = 0.75 Batch timing = 10 Hrs Cost per unit = 4.12 Rs.
22 KW * 0.75 = 16.5 KW
Batch Timing = 10 Hr
16.5*10 = 165 KW/Hr
Cost = 165 * 4.12
Cost = 679.8 Rs.
CACL2 DEHYDRATION Motor capacity = 22 KW Motor error = 0.75 Batch timing = 10 Hrs Cost per unit = 4.12 Rs.
22 KW * 0.75 = 16.5 KW
Batch Timing = 10 Hr
16.5*10 = 165 KW/Hr
Cost = 165 * 4.12
Cost = 679.8 Rs.
NAOH DEHYDRATION Motor capacity = 22 KW Motor error = 0.75 Batch timing = 15 Hrs Cost per unit = 4.12 Rs.
22 KW * 0.75 = 16.5 KW
Batch Timing = 15 Hr
16.5*15 = 247.5 KW/Hr
Cost = 247.5 * 4.12
Cost = 1019.7 Rs.
COST ESTIMATION
STUDY The cost is estimated by evaluating the cost
of steam, cooling water and power consumed by the agitator.
The cost incurred by every process is studied.
The cost of all the process is evaluated to find the money spent for the entire recovery process per batch.
TDS REMOVAL Mass of steam = 1658.8 Kg Cold water (condensor 1) = 238.08 Kg Cold water (condensor 2) = 29.89 Kg
Cost of steam = 1658.8 * 1.160.6 = 1925.2032 RsCold water (condensor 1) = 238.08 * 9.6 = 2285.65
Rs.Cold water (condensor 2) = 29.89 * 0.96 = 28.694 Rs.
Power consumed = 679.8 Rs.
Total cost = 4919.3472 Rs.
EA & THF ENRICHMENT Mass of steam = 1260 Kg Cold water (condensor 1) = 180.10 Kg Cold water (condensor 2) = 19.67 Kg
Cost of steam = 1260 * 1.1606 = 1496.4 RsCold water (condensor 1) = 180.10 * 9.6 =
1728.96 Rs.Cold water (condensor 2) = 19.67 * 0.96 =
18.8832 Rs.
Total cost = 3244.2432 Rs.
EA ENRICHMENT
CACL2 DEHYDRATION
Power consumption cost= 679.8 Rs.
Total Cost = 679.8 Rs.
THF REMOVAL Mass of steam = 1125 Kg Cold water (condensor 1) = 161.9 Kg Cold water (condensor 2) = 15.94 Kg
Cost of steam = 1125 * 1.1606 = 1305.675 RsCold water (condensor 1) = 161.9 * 9.6 =
1554.24 Rs.Cold water (condensor 2) = 15.94 * 0.96 =
15.3024 Rs.
Total cost = 2875.22 Rs.
THF ENRICHMENT
EA REMOVAL Mass of steam = 1250 Kg Cold water (condensor 1) = 189.47 Kg Cold water (condensor 2) = 20.94 Kg
Cost of steam = 1250 * 1.1606 = 1450.75 RsCold water (condensor 1) = 189.47 * 9.6 =
1818.912 Rs.Cold water (condensor 2) = 20.94 * 0.96 =
20.1024 Rs.
Total cost = 3289.76 Rs.
NAOH DEHYDRATIONPower consumption Cost = 1019.7 Rs.
Total Cost = 1019.7 Rs.
ANHYDROUS THF RECOVERY
Mass of steam = 1120 Kg Cold water (condensor 1) = 174.17 Kg Cold water (condensor 2) = 19.41 Kg
Cost of steam = 1120 * 1.1606 = 1299.87 Rs.Cold water (condensor 1) = 174.17 * 9.6 =
1672.032 Rs.Cold water (condensor 2) = 19.41 * 0.96 =
19.6336 Rs.
Total cost = 2991.536 Rs.
TOTAL COSTPROCESS COST (Rs.)
FLASHING 4919.3472
EA & THF ENRICHMENT 3244.2432
CACL2 DEHYDRATION 679.8
THF REMOVAL 2875.22
EA REMOVAL 3289.76
NAOH DEHDYRATION 1019.7
ANHYDROUS THF REC. 2991.536
TOTAL 19019.6058
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