I would like to design a solar hydrogen production system using PV solar cells to drive Alkaline Water Electrolyser, which requires studying the productivity of the water electrolyzer to estimate the amount of water needed.
If you have the water quantity say 0ne grams, then you can first calculate the number of H2 molecules contained in the one gram mole of water= Avogadros number = one gram mole of hydrogen = 2.016 g/mol,
one gram mole of water= 18.01528 grams
So the ratio between H2 and water is
H2:water= 1/9
So, you can easily calculate either quantities if you have the other.
To produce one kg of H2 you need 9 kg of water.
To get one molecule of hydrogen you need two electronic charges 2q
The charge Q required to get one kilograms= 2qx no. of moles in one kilograms
= 2q x (1000/ 2.016 )x Avogadro number
with q= 1,6x10^-19 As and Avogadro number= 6.0221409e+23
Also every molecule needs a voltage to split it say Vsplit
The electrical energy required to produce one kilogram hydrogen= Q x Vsplit.
Vsplit minimum = 1.229 V
Practically one needs more than 1,229 V to split the water and the overvoltage required causes a conversion efficiency less than one.
If you have the water quantity say 0ne grams, then you can first calculate the number of H2 molecules contained in the one gram mole of water= Avogadros number = one gram mole of hydrogen = 2.016 g/mol,
one gram mole of water= 18.01528 grams
So the ratio between H2 and water is
H2:water= 1/9
So, you can easily calculate either quantities if you have the other.
To produce one kg of H2 you need 9 kg of water.
To get one molecule of hydrogen you need two electronic charges 2q
The charge Q required to get one kilograms= 2qx no. of moles in one kilograms
= 2q x (1000/ 2.016 )x Avogadro number
with q= 1,6x10^-19 As and Avogadro number= 6.0221409e+23
Also every molecule needs a voltage to split it say Vsplit
The electrical energy required to produce one kilogram hydrogen= Q x Vsplit.
Vsplit minimum = 1.229 V
Practically one needs more than 1,229 V to split the water and the overvoltage required causes a conversion efficiency less than one.