Follow the units.
Assumptions:
1. Location in area with solar energy/ solar insolation of 5kWh/m^2/day (midwest-ish).
2. 10% solar absorption efficiency, 75% electrolysis conversion of the 10% = 7.5% solar to chemical
3. standard temp and pressure
1Wh=3600Ws=3600J (Unit equivalences)
(7.5%)(5000Wh)=375Wh
(375Wh)(3600s/h)=1,350,000J
Reaction at 1.5V
1,350,000J/1.5V=900,000C
9,00,000C/(1.602E-19)[C/electron]=5.6174E+24 electrons
5.6174E+24[electrons]/6.022E+23[mol^-1]= 9.3281E+00[mol] of electrons
Taking note of chemical equ: (1*O2 + 2*H2 = 2*H2O + 4 electrons)
The 2H of the water (H2O) each need 1 electron, so if we have 9.328 mol of electrons, only 4.66 mol of water would be split.
This results in 4.66 mol (104.5L)of H2 and 2.33 mol (52.23L)of O2 that got created from that m^2 over the course of the day.
That is also 4.66 mol of water split per day from a 1 square meter.
Molar mass of H2O is 18.015g/mol.
1 liter of water is about 1000g
3.785L = 1 gal H2O
3785g = 1 gal H2O
(4.66mol)(18g/mol) = 84g (0.0222gal)of H2O is converted in the m^2 per day.
4047 m^2 = 1 acre
4047 [m^2]*84g = 339,948g or about 340Kg of H2O converted per acre per day
340,000[g] / 3785[g/gal] = approx 90 gal of water per day per acre
4.66mol of H2 per m^2 per day; 4.66*4047=18,859mol (422,442 L)of H2 at STP.
Think of a 70L (18.5gal) tank at 10,000 PSI (700Bar or approx 690atm) on a fuel cell car.
P1V1=P2V2
(422,442 L)(1atm)/(690atm) = 612L of H2 at 690atmospheres (10,000psi)
612L@10,000psi / 70[L/ave tank] = 8.74 (70L tanks)
8.74 tankfuls of hydrogen from 1 acre per day in moderate solar exposure from a net 7.5% solar to chemical efficiency. Reduce this by 20% for compressing.
Long time trader of HYSR and now an accumulator. This and other analysis works for me. I know enough science and semiconductor engineering where BS science doesn't work on me. It may help when it keeps the price of HYSR down.
