Electrochemical Engineering, Chlor-Alkali & Industrial Electrolyzers
Chlor-Alkali Membrane Cell NaOH, Cl2 ECU & kWh/Ton Formula
Calculate Daily 100%-Basis Caustic Soda NaOH Production (t/d), Co-Product Chlorine Gas Cl2 (t/d at 0.886 ECU Ratio), Rectifier DC Load (MW), and Specific DC Energy (kWh/t NaOH).
Every 1.00 ton of dry NaOH electrolyzed co-produces an inescapable 0.886 tons of Cl2—the Electrochemical Unit (ECU)!
Ion-exchange membrane chlor-alkali cellrooms electrolyze saturated purified NaCl brine (2NaCl + 2H2O -> Cl2 + H2 + 2NaOH): passing 1 mole of electrons (Faraday constant F = 96,485 C/mol) forms 1.0 mole of NaOH (MW = 40.00 g/mol) at the cathode and 0.5 mole of Cl2 gas (MW = 70.906 g/mol -> 35.453 g Cl2 per 40.00 g NaOH = 0.8863 ECU ratio!) at the anode! Across N_cells series-connected cells carrying DC current I (kA) at cathodic current efficiency eta_F (typically 95%–97% due to minor OH- back-migration across the perfluorosulfonic/carboxylic membrane), Daily Dry NaOH Output is (I_kA * 1,000 * N_cells * 86,400 * eta_F / 96,485) * 40 / 10^6 t/d, and Specific DC Energy Consumption reduces directly to (V_cell * 96,485) / (3.6 * 40 * eta_F) = (670.035 * V_cell) / eta_F kWh per metric ton NaOH!
Verify on a 100-Cell Bipolar Membrane Electrolyzer—Cell_Current I = 18.0 kA (B2), Bipolar_Cell_Count N = 100 (B3), Current_Efficiency eta_F = 0.96 (96.0%, B4), and Operating_Cell_Voltage V_cell = 3.10 V (B5): Daily Dry 100% NaOH Output hits 61.90 t/d (E2), Co-Product Chlorine Cl2 equals 54.86 t/d (E3), Cellroom DC Power Load equals (18 * 100 * 3.10) / 1,000 = 5.58 MW (E4), and Specific DC Power Consumption equals 2,163.65 kWh/t NaOH (E5)!
| Chlor-Alkali Cellroom KPI (18 kA, 100 Cells, 96% Eff, 3.10 V/cell) | Formula | Verified Output |
|---|---|---|
| Dry 100% NaOH Produced (t/d) | (I_A * N * 86400 * eta / 96485) * 40 / 1e6 | 61.90 t/d |
| Co-Product Chlorine Cl2 (t/d) | (I_A * N * 86400 * eta / 96485) * 35.453 / 1e6 | 54.86 t/d (0.886 ECU ratio) |
| Cellroom DC Rectifier Load (MW) | (Current_kA * N_Cells * Cell_V) / 1000 | 5.58 MW (6.12 MW @ 3.40 V) |
| Specific DC Energy (kWh/t NaOH) | (Current_kA * N * Cell_V * 24) / NaOH_t_d | 2,163.65 kWh/t (2,373.04 @ 3.40 V) |
Faradaic Membrane Chlor-Alkali Specific Energy & NaOH Tonnage Formula
Validate Current_kA > 0, integer Cells >= 1, Current_Efficiency in (0.75, 1.0], and Cell_Voltage >= 2.20 V (above reversible decomposition voltage ~2.19 V)!
=LET(ik_a, B2, nc_l, B3, ef_f, B4, vc_l, B5, IF(OR(ik_a <= 0, nc_l < 1, INT(nc_l) <> nc_l, ef_f <= 0.75, ef_f > 1, vc_l < 2.2), "Invalid Electrolyzer Input", LET(ml_d, (ik_a * 1000 * nc_l * 86400 * ef_f) / 96485, na_t, (ml_d * 40) / 1000000, ROUND((ik_a * nc_l * vc_l * 24) / na_t, 2))))Aged Membrane Voltage Creep Energy Penalty Check (Set Cell Voltage = 3.40 V)
Simulate 4-year membrane scaling/cathode-coating aging raising Cell_Voltage B5 from 3.10 V to 3.40 V: Daily NaOH Output E2 stays fixed at 61.90 t/d (governed purely by Amps and eta_F!), while DC Load E4 climbs to 6.12 MW and Specific Energy E5 jumps by +209.39 kWh/t to 2,373.04 kWh/t!
=IF(ISTEXT(E5), E5, ROUND((B2 * B3 * B5) / 1000, 2))Try the example
Use the original values or change them to test your understanding.