Water Electrolysis Simulation
Theory: current sets hydrogen, voltage sets power
Overview
What should I look at when comparing electrolyser performance?
Split it in two. The amount of hydrogen depends only on current: each molecule needs two electrons, so 1 kA gives 37.6 g per hour regardless of electrolyser type. The voltage needed to drive that current sets the power used: theoretical voltage plus activation loss to drive electrode reactions and ohmic loss in membrane and parts. So performance is compared by how low a voltage a cell needs at a given current density — the polarization curve. Raising current density makes more hydrogen from the same electrode area but raises voltage and electricity per kg. That trade-off is at the heart of design.
Cell-voltage model
E_rev is theoretical voltage (1.18 V at 80 °C), b the Tafel slope (0.030 V), i₀ exchange current density (10⁻⁶ A/cm², governed mainly by the oxygen-evolving anode) and R_ASR area-specific resistance (0.15 Ω·cm²). Unlike a fuel cell, losses add to the voltage.
Hydrogen output, specific energy and heat
F is the Faraday constant and M the molar mass of hydrogen (2.016 g/mol). 1.48 V is hydrogen's higher heating value (39.4 kWh/kg) expressed as voltage; voltage above it becomes heat q.
The same year as the battery
Nicholson and Carlisle electrolysed water in 1800, just after Volta announced his battery: they noticed bubbles from water connected to the battery and identified them as hydrogen and oxygen. In the 20th century, cheap hydropower drove large alkaline electrolysers producing hydrogen for ammonia. Making hydrogen from natural gas later became cheaper and electrolysis lost its lead, but large plants are being built again as a way to make hydrogen from renewable electricity.
Worked examples
Example 1: voltage, energy and heat by current density (PEM, 80 °C)
| Current density | Cell voltage | Activation | Ohmic | Energy per kg H₂ | Efficiency (HHV) | Heat |
|---|---|---|---|---|---|---|
| 0.5 A/cm² | 1.649 V | 0.394 V | 0.075 V | 43.8 kWh | 90% | 0.084 W/cm² |
| 1.0 A/cm² | 1.744 V | 0.414 V | 0.150 V | 46.4 kWh | 85% | 0.264 W/cm² |
| 2.0 A/cm² | 1.915 V | 0.435 V | 0.300 V | 50.9 kWh | 77% | 0.871 W/cm² |
| 3.0 A/cm² | 2.077 V | 0.447 V | 0.450 V | 55.2 kWh | 71% | 1.792 W/cm² |
Raising current density from 0.5 to 3.0 A/cm² makes six times the hydrogen from the same area, but energy per kg rises from 43.8 to 55.2 kWh, about 26%. Activation loss barely changes; the increase is almost all ohmic. Voltage reaches 1.48 V at about 0.02 A/cm²; above that the cell releases heat.
Example 2: a 10 MW electrolyser
| Item | Value |
|---|---|
| Hydrogen per kA | 37.6 g/h（0.418 Nm³/h） |
| Operating point | 2.0 A/cm², 1.915 V |
| Hydrogen output | about 196 kg/h |
| Electrode area needed | about 261 m² |
| Heat | about 2.3 MW |
Heat is 0.871 W/cm² × 261 m², about 2.3 MW. These energies are for the cells alone; a real plant adds power for rectifiers, pumps, hydrogen drying and other auxiliaries.
If lower current density means higher efficiency, why not always run low?
It is a balance between electricity cost and equipment cost. At 0.5 A/cm² the energy per kg is about 14% less than at 2 A/cm², but the same hydrogen output needs four times the electrode area — that many more cells, membranes, platinum-group catalyst and stainless or titanium parts. With expensive power and near-continuous running, efficiency wins and current density goes down; with cheap but short-lived power such as surplus solar, a smaller electrolyser at higher current density gives lower cost per kg. Higher current density also means more heat, uneven membrane temperature and damage starting at hot spots, so cell temperature distribution is coupled with membrane degradation to see how far current density can be pushed.
Modelling workflow
- Fit exchange current density and resistance to measured polarization curves.
- Solve water and gas-bubble flow in channels and porous layers.
- Reflect bubble coverage of the electrode in active area.
- Derive heat from current distribution and couple temperature with membrane resistance.
- Simulate operation under fluctuating power over time and estimate degradation.
“An electrolyser that degraded with every shutdown”
In a solar-powered electrolyser, voltage rose faster than in steady operation. Analysis alongside operating logs showed current dropping to zero at every sunset, swinging electrode potential and gradually damaging catalyst and component surfaces. Restarting at full current while still cold also meant high membrane resistance and strong local heating. Keeping a small current during shutdown and limiting current at start-up until the cell warmed slowed the voltage rise.
Common mistakes
Mistakes and fixes
| Mistake | Effect | Fix |
|---|---|---|
| Using 1.23 V at 25 °C | Temperature effect wrong | Value at operating temperature |
| Mixing LHV and HHV efficiency | Efficiency off by about 15% | State the basis |
| Ignoring bubbles | Voltage underestimated at high current | Include bubble coverage |
| Cell energy only | Plant energy underestimated | Add auxiliaries |
| Lifetime from steady operation only | Faster degradation with variable power | Include start-stop |
I'd like to learn related topics.
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