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Stan’s Legacy

Coefficient of performance

How much energy is in the gas produced, against the electrical energy it took?

The formula
E=V·hvW=P·tC=EW
C
Coefficient of performance
E
Chemical energy out
W
Electrical energy in
V
Hydrogen collected
P
Electrical power in
t
Duration, s
LaTeX 
E = V \cdot h_v \qquad W = P \cdot t \qquad C = \frac{E}{W}

Work it out

Millilitres of hydrogen — not of the mixed gas. Oxygen carries no fuel value.

Watts, measured as true power — the mean of v(t)·i(t), not the product of two meter readings.

How long that power was drawn.

Which figure to value the hydrogen at.

Method

  1. Take the hydrogen volume alone. The oxygen produced alongside it carries no fuel value, so valuing the mixed gas overstates the output by half.
  2. Multiply by the heating value: 10.783 kJ per litre for the lower figure, 12.745 for the higher. Use the lower unless the exhaust is condensed and that latent heat is actually recovered — in a flame open to the room, it is not.
  3. Multiply the electrical power by the run time for the energy in. The power must be true power: the mean of instantaneous voltage times instantaneous current. On a pulsed or resonant waveform, multiplying an RMS voltmeter reading by an RMS ammeter reading gives apparent power, which can be several times the real figure.
  4. Divide energy out by energy in.
  5. Also report kilojoules per litre. It is a harder number to fool, comparable across rigs, and commercial electrolysers sit around 12 to 18 kJ per litre.

Assumptions

  • The input power is true power. This is where most reported figures above unity come apart: on a pulsed waveform, voltage and current are not in phase, and the product of two meter readings is apparent power rather than real power.
  • The volume is dry hydrogen at the stated conditions. Gas collected over water carries vapour; mixed gas counted as hydrogen inflates the output by 50 %.
  • The energy in the gas is chemical energy released on burning it, not work recovered. An engine or fuel cell then returns a fraction of it — thirty to sixty per cent — so a coefficient near 1 here is not a self-sustaining system.
  • Only the electrolysis power is counted. A complete accounting includes the driver, the pump and any heating, which is why "at the wall" and "at the cell" can differ severalfold.
  • The archive takes no position on whether this ratio can exceed one. It takes a position on the arithmetic being visible, which is what this page is for.