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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
PROCESS FLOW DIAGRAM
Membrane Cell Process
MATERIAL BALANCE
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).
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.
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
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
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.
ENERGY BALANCE
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
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
• 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
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
• 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
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
• 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
% 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 %
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)

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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
  • 3.
  • 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.
  • 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)