Electrolysis Question [RE-wrenches]

[Antony Tersol]

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My mistake.  The google search target expressed it as 237 kJ/mole.
I converted the kJ/mole to kWh/kg.

Since it is per mole of hydrogen GAS, 1 mole weighs 2 grams (H2 for the gas), not 1 gram as I used (H+ ion).

therefore 237 kJ electrical energy per mole = .0658 kWh / 2 g = 32.9 kWh / kg.





-------Original Message-------

 

From: RE-wrenches at topica.com

Date: Wednesday, March 31, 2004 10:00:54 AM

To: RE-wrenches at topica.com

Subject: re: Electrolysis Question [RE-wrenches]



Not having an extensive background in chemistry I am not sure why google
search came up with this number but I would assume it has something to do
with monatomic vs diatomic since this equation is exactly half of actual. If
this is not so then my electrolyzers are in the range of 130% efficient
since my data logger conforms with 32.9 kwh's/kg x 1.34 just as posted on
DOE ,NREL and many others. Also Stuart and Proton Energy and others list
their less than %100 efficient electrolyzers at 48 to 50 kwh's/kg. In
addition efficiency is dependent on cell temperatures so this may be a
factor in this equation.

 



 

237 kJ electrical energy per mole = .0658 kWh / mole = .0658 kWh / g =

65.8 kWh / kg

 

1 mole = 6.023 x 10 ^23 molecules

1 amu = 1.659 x 10^-27 kg

1 mole of hydrogen weighs 9.9 x 10^-4 kg = 1 g.

1 J = 1 watt-sec

1kJ = 1 kW-sec = 1 kWh/3600


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