MANUFACTURE OF CHLORINE - CAUSTIC SODA USING ELECTROLYSIS PROCESS (MEMBRANE CELL) * CORRECTION NEEDED
Download as PPT, PDF27 likes16,822 views
This document summarizes the process of manufacturing chlorine and caustic soda using electrolysis. It includes: - A process flow diagram of the membrane cell process used to separate NaCl into NaOH, H2, and Cl2 via electrolysis. - Material and energy balances calculations for each unit operation including the membrane cell, evaporator, and dryer. These calculate chemical reactions, flows, heating needs and efficiencies. - The process achieves 70.28% conversion of NaCl and 27.37% yield of NaOH from the reacted NaCl. Waste streams and energy requirements are also quantified.
MANUFACTURE OF CHLORINE - CAUSTIC SODA USING ELECTROLYSIS PROCESS (MEMBRANE CELL) * CORRECTION NEEDED
- 1. PROJECT TOPIC MANUFACTURE OF CHLORINE– CAUSTIC SODA USING ELECTROLYSIS PROCESS BHARATI VIDYAPEETH COLLEGE OF ENGINEERING CHEMICAL DEPARTMENT Prof. R.K.Kulkarni Presented by Ankush Gupta
- 2. PROCESS FLOW DIAGRAM Membrane Cell Process
- 5. Basis : 1150 kg/hr NaOH , 1000 kg/hr Chlorine 2NaCl + 2H2O 2NaOH + H2 + Cl2 (58.5) (18) (40) (2) (70) (28.75) (28.75) (28.75) (14.375) (14.375) Kmoles/hr (1681.8) (517.5) (1150) (28.75) (1006) kg/hr • Sample calculation Moles = weight / molecular weight For NaCl Moles = 1150/40 28.75 Kmol/hr NaCl required (Kg) = 28.75 * 58.5 1681.8 kg (1700 Kg/hr).
- 6. Assumption : 1700 kg/hr of NaCl 20 kg/hr Na2Co3 • 15/100 x Feed =1700 Feed = 11333.33 Kg/hr. OVERALL MATERIAL BALANCE Feed (F) + Na2Co3 = |Sludge|4 + |NaCl|3 11333.33 + 20 = |Sludge|4+ |NaCl|3………………………eqn(1) MATERIAL BALANCE OF NaCl 1700 = 28%|NaCl|3 + 10%|Sludge|4 170000 = 28|NaCl|3 + 10|Sludge|4…………………………….eqn(2) On Solving Eqn(1) & eqn(2) we get |NaCl|3 = 3137.03 Kg/hr. |Sludge|4 = 8216.3 Kg/hr.
- 7. For 55% efficiency of Caustic Soda &Cl2: 2NaCl + 2H2O = 2NaOH + H2 + Cl2 Mass flow(kg/hr) 878.36 270.25 330.28 8.256 289.01 Molecular Weight 58.5 18 40 2 70 Moles flow(Kmol/hr) 15.014 15.014 15.104 7.552 7.552 Moles flow(Kmol/hr) 8.257 8.257 8.257 4.153 4.153 (55%) Sample Calculation: For NaOH : • Amount of NaCl entering a membrane cell = 28%|NaCl|3=28% (3137.03)=878.368kg/hr Converting into molar flow rate = 878.368/58.5=15.014 Kmol/hr •By Stoichiometry , No of moles of NaCl = No of moles of NaOH = 15.014 Kmol/hr. But since we have for 55% efficiency for Membrane cell • No of moles of NaOH = 55%(15.014)= 8.257 Kmol/hr. Therefore the amount of NaOH obtained= 8.257*40=330.28 Kg/hr
- 8. OVERALL MATERIAL BALANCE OVER EVAPORATOR Aq. NaOH(35%NaOH) = Aq.NaOH(50% NaOH) + Water vapour 330.28 = 231.196 + 99.081 Calculation for amount of aq.NAOH obtained as a main product: • Pure NaOH out of aq. NaOH = 35% of (330.28) = 115.598 kg/hr • Water content in incoming aq. NaOH = 330.28 – 115.98 = 214.68 kg/hr Amount of NaOH obtained after 50% concentration is • 50% of (aq. NaOH) = 115.598 kg/hr • Therefore aq. NaOH (50%) obtained as product = 231.196 kg/hr • Water content in obtained aq. NaOH (50%) = 231.196-115.598=115.598 kg/hr • Therefore Water Vapour = 214.68-115.98=99.082 kg/hr
- 9. OVERALL MATERIAL BALANCE OVER DRYER: Wet Cl2 + H2SO4(98% conc) = Dry Cl2 + H2SO4 (70% conc) 289.019 + 98% of (25kg) = Dry Cl2 + H2SO4 (70%) …….eqn1 MATERIAL BALANCE OF H2SO4 OVER DRYER: Inlet H2SO4 (98%conc) = Outlet H2SO4 (70%conc) 98%(25) = 70% H2SO4 98x25 = 70% H2SO4 H2SO4 (70%) = 35 kg/hr Substituting in …..eq1 Dry Cl2 = 278.519 kg/hr.
- 10. ENERGY BALANCE
- 11. STEAM REQUIREMENT FOR HEATING BRINE m = mass flow rate of Aq NaCl in kg/hr CP= specific heat capacity of Aq NaCl At 60O C ΔT = Temperature Difference λ = latent Heat of evaporization in kJ/kg m1 = Flow rate of steam kJ/hr mcPΔT = m1 λ 3137.03 x 3.274 (333-298) = m x 2358.40 m1 = 151.186 kJ/hr
- 12. ENERGY BALANCE OVER MEMBRANE CELL Assumption Datum temperature = 25o C INLET STREAM OUTLET STREAM MATERI AL NAME SPECIFIC HEAT AT 60o C FLOW RATES kg/hr H2O 4.185 kJ/kgK 270.252 Aq.NacL solution 3.247 kJ/kgK 876.368 MATERIAL NAME SPECIFIC HEAT AT 80o C FLOW RATES kg/hr Hydrogen (g) 14.43 kJ/kgK 8.256 Chlorine (l) 0.48 kJ/kgK 289.019 NaoH (35% by wt) 3.594 kJ/kgK 330.28 Depleted brine 3.247 kJ/kgK 521.065
- 13. • INPUT STREAM Material mcPΔT Aq .NaCl 876.368 x 3.247x (333-298) = 99594.84 kJ/hr H2O 270.252 x 4.185 x (333-298) = 39585.16 kJ/hr TOTAL HEAT INPUT = 99594.84 + 39585.16 = 139180 kJ/hr • OUTPUT STREAM Material mcPΔT Hydrogen (g) 8.256 x 14.43 x (353-298) = 6552.37 kJ/hr Chlorine (l) 0.48 x 289.019 x(353-298) = 4855.51 kJ/hr Aq.NaoH 3.594 x 330. 28 x (353-298) = 41545.92 kJ/hr Depleted Brine 3.247 x 521. 065 x (353-298)= 59216.43 kJ/hr TOTAL HEAT OUTPUT = 6552.37 + 4855.51 + 41545.92 + 59216.43 = 112170.23 kJ/hr 2Na (s) + 2H2O 2NaOH + H2O ΔH = -368.4 kJ/gmNa Depleted brine = 526.064 kg/hr (50% by wt of NaCl) 0.5 x 526.064 x -368.4 = - 96900.98 J/hr
- 14. MATERIAL NAME SPECIFIC HEAT AT 80o C FLOW RATES kg/hr Aq.NaoH 35% by wt 3.594 kJ/kgK 330.28 MATERIAL NAME SPECIFIC HEAT AT 110o C FLOW RATES kg/hr 50 % Caustic soda 3.564 kJ/kgK 231.196 INLET STREAM OUTLET STREAM MATERIAL NAME Latent Heat AT 110o C FLOW RATES kg/hr Steam 2231.86 kJ/kg 94.127 ENERGY BALANCE OVER EVAPORATOR
- 15. • INPUT STREAM Material mcPΔT Aq .NAOH 330.28 x 3.594 x (353-298) = 65286.44 kJ/hr TOTAL HEAT INPUT = 65286.44 kJ/hr • OUTPUT STREAM Material mcPΔT Aq.NaoH (50% by wt) 231.196 x 3.564 x (383-298) = 45319.03 kJ/hr Material mλ Water vapour (g) 94.127 x 2231.86 = 210078.28 kJ/hr TOTAL HEAT OUTPUT = 45319.03 + 210078.28 = 255397.31 kJ/hr
- 16. MATERIAL NAME SPECIFIC HEAT FLOW RATES kg/hr Wet chlorine 0.48 kJ/kgK 289.019 H2SO4 (98% by wt) 1.465 kJ/kgK 10 MATERIAL NAME SPECIFIC HEAT FLOW RATES kg/hr Dry chlorine 0.48 kJ/kgK 285.019 H2SO4 (70% by wt) 2.177 kJ/kgK 14 INLET STREAM OUTLET STREAM ENERGY BALANCE OVER DRYER
- 17. • INPUT STREAM Material mcPΔT Wet Chlorine 289.019 x 0.48 x (353-298) = 7630.10kJ/hr TOTAL HEAT INPUT = 7630.10 kJ/hr • OUTPUT STREAM Material mcPΔT Dry Chlorine (l) 285.019 x 0.48 x (343-298) = 6156.41 kJ/hr H2SO4 70% 14 x 2.177 x (303-298) = 152.39 kJ/hr TOTAL HEAT OUTPUT = 6156.41 + 152.39 = 6308.8 kJ/hr
- 18. % conversion NaCl = (Reacted moles / Total no moles) x 100 = {(878.36 – 261.032)/878.36} x 100 = 70.28 % %Yield = (NaoH moles Formed /Reacted moles of NaCl) x 100 = {(115.598/ 40) / (617.328/58.5) } x 100 = 27.37 %
- 19. REFERENCES 1 ) Subrata Basu , Swapan Kumar Mukhopadhyay , Amitava Gangopadhyay and Sujata G. Dastidar “International Research Journal of Environment Sciences Characteristic Change of Effluent from a Chlor-alkali Industry of India due to Process Modification” 2013. 2) S.Koter.A.Warszawski “Polish Journal of Environmental Studies Electro- membrane Processes in Environment Protection” (2000) 3) Yohannes Kiros and Martin Bursell “International Journal Electrochemistry Science Low Energy Consumption in Chlor-alkali Cells Using Oxygen Reduction Electrodes (2008)” 4) Rezaee, J. Derayat, S.B. Mortazavi,Y. Yamini and M.T. Jafarzadeh “American Journal of Environmental Sciences Removal of Mercury from Chlor-alkali Industry Wastewater using Acetobacter xylinum Cellulose(2005)” 5) Dryden’s outlines of chemical technology (2012)