I am using a DC power which is connected to the external main power with max voltage output of 31V and max current output of 5A to run a Hoffman like electrolyzer.
In many textbooks, many have experimented with electrolyzer to produce hydrogen have used as low as 1-2V and still getting a high current.
I get very low efficiency because of high voltage as I use this formula: efficiency= HHV* hydrogen volume ( ml) / V*I
where HHV is high heating value, V is voltage and I is current.
Maybe I should only consider a portion of the DC power voltage as electrolyzer might have a max limit for voltage?
Hi Javad,
Can you clarify the following items:
*Number of the cells
*Active area of the electrolyser
*Are you measuring both current and voltage of the electrolyser?
*What value are you using for HHV and what is the unit?
Cheers,
Reza
I agree with Reza Omrani . I hope that you use units that after application this HHV* hydrogen volume ( ml) you get Watts. And for one cell you need 1.7-2 V so with 31 V you will get very poor efficiency. If you are using Hoffman electrolyser, do you use Pt electrodes? If yes, voltage can be arroud 2 V because you are using Pt catalyst. If not you need higher voltage .. with some proper metal or carbon electrode 10 V should be enough. I would add one more question. Which electrolyte do you use ?
Reza Omrani
Number of cells: Using two platinum electrodes as cathode and anode each with a surface area of 0.45cm^2.
The DC power source is held constant at 31V and thus only current varies.
I am using 12.7MJ/m^3 which gives 12.7J/milli liter.
Roman Fiala
Yes the electrodes are platinum plates each with area of 0.45cm^2 and the electrolyte i am using is sodium sulfate. Weight fraction of sodium sulfate is varied between 5-15wt.%.
1.7-2V for pt electrodes gives me a good clue on efficiency. So maybe even though my DC power source is 31V, actually only 1.7-2V is taken up by the electrodes?
One more thing: When i measure current versus voltage, the minimum voltage is required to be around 3V based on my power source so this might be the same is the reversible voltage which is around 1.23V. For example: at 3V current is 0.05A, at 20V current is 0.12A and at 31V current is 0.250A for a solution containing 10wt.% of sodium sulfate.
Dear Javad Hikmati
This I do not understand: "So maybe even though my DC power source is 31V, actually only 1.7-2V is taken up by the electrodes?" Voltage on the elctrodes is that one that you meassure by voltmeter.
And the another thing I did not understand also. If your power source cannot apply lower voltage than 3 V, use another power source. Power for efficiency formula is always voltage on the electrodes (measured by voltmeter directly on the electrodes for sure) multiplied by current. Of course that you can use higher voltage and you get higher power but you would get also lower efficiency. I only wanted to say that with using Pt electrodes you can use voltage less than 2 V and get good efficiency. Without Pt electrodes you need higher voltage (around 10 V and efficiency is low.)
Roman Fiala
I have not used voltmeter and i will cheack with my supervisor if we can find one in the laboratory. The only thing that is actually varies is current which is shown in the display of the DC source and the voltage on that display is shown to be 31.
The voltage on thr display is maybe ok but I think that display on zhe power source is not enough accurate.
Roman Fiala Thanks.
I think you are right about the voltage of platinum electrodes.
The DC power source has a voltage between 0-31V and 0-5Ampere.
During the operation the voltage I use the maximum voltage which is displayed 31V on the power source and only current varies during the electrolyses. For example current goes up from 100mA to 300mA if the temperature of electrolyte is increased.
I have heard with my supervisor and we may be able to order a voltmeter or even a multimeter that is able to measure voltage, current and resistance when connected to the electrodes.
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