From the above information can we calculate the Elektron MMW?

[BoyntonStu]

According to this document, the Mir electrolysis system had a mass of 140 kg, had twelve cells each with a cell voltage of 2v and a size of 0.8m x 0.13m, used 800 w with max current at 2V of 400 amp.


80 cm x 13 cm x 12 cells = 12480 cm^2 Area of all cells

12480 cm^2 = 1934 in^2

400 A / 1934 in^2 = .21 A/in^2

Nice to know that we are in the same ballpark with NASA..

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Oxygen is generated in the system by the electrolysis of the water content in potassium hydroxide solution. Water is made up of 89% oxygen by weight. The reaction breaks down water into its constituents of oxygen and hydrogen. The oxygen is released directly into the atmosphere of the SM and the hydrogen is vented to vacuum. The electrolysis unit is made up of 12 electrolysis cells which are enclosed in a blastproof housing. The cells are cooled by the thermal control system (STR). The decomposition of 1 kg (2.2 lbs) of water yields 25 L (0.88 ft3) of oxygen per hour at a pressure of 760 mmHg, which is enough to support one crew member for one day. To provide the daily amount of oxygen for 3 - 4 crew members, 3 - 4 kg (6.6 - 8.8 lbs) of water must be decomposed. Power consumption of the process is ~ 1 kW.

From the above information can we calculate the Elektron MMW?

BoyntonStu

[Painless]

Hmm, lets see:

Volts: 12 cells @ 2v per cell = 24v total

Amps: 400

ML produced: 25,000 (25L/hr = 0.4167 LPM)

Production time for above: 3600 seconds

I make that 0.04 MMW according to the calculator at http://www.watervan.co.uk/tools.php

However, this only takes into account the oxygen. If we are making 25L of oxygen an hour, how much hydrogen would be produced?


According to this page:

http://woelen.scheikunde.net/science/chem/exps/precision_electrolysis/index.html

...H2 and O2 are produced from electrolysis in an exact 2:1 volume ratio, therefore, if we produce 25L of O2 we have 50L of H2, which equals 75L of HHO...

So, back to our calculations:

Volts: 12 cells @ 2v per cell = 24v total

Amps: 400

ML produced: 75,000 (75L/hr = 1.25 LPM)

Production time for above: 3600 seconds

Therefore, the MMW is 0.13 ... ?

Anyone concur?

[Painless]

Thinking about this further, it's also possible that the cells are wired in parallel?

If that were true, our calcs would look like this:

Volts: 12 cells in parallel @ 2v per cell = 2v total

Amps: 400

ML produced: 75,000 (75L/hr = 1.25 LPM)

Production time for above: 3600 seconds

This gives us 1.56 MMW