hmmmm i always figured injection time changes when revolution time changes (rpm) and the changes are made proportionately with rpm.. having a predetermined variable being horsepower.
seb check my math above once more please.. pretty sure i have it right... so it is 7.4 ml per cylinder for 50 hp motor amounting to 2.6 liters per hour at 65mph and 2000rpm? and a average hydrocarbon fuel gets around 6.5 liters a hour at 65mph and 2000rpm due to the fact that there is 2.5 more power in water from having a higher quantity of hydrogen per 1:1 by volume.
Now you corrected it right? Attention because you still confusing micro liters with milliliters. But now the results seems right! I don't understand why you talk about 2000 rpm? Do you mean for 6 cylinders? still not clear...
however
stan said 7.4 micro liters per injection cycle for a 50hp 4 cylinders running at 3000rpm and 65mph
I believe he used the mixing of the gas to control the motor speed i mean he used more or less exhaust recirculation accordingly with the rpm desired. I say this because he talked about the speed of the explosion of the hydrogen witch would be to great alone. So to reduce the speed or the rpm (MEYER TRANSLATION) he substituted the air in for exhaust gas. why? Because is much easier (KISS) to make a pulse of fix width that changes only in frequency (rpm) than is to create a pulse that changes frequency and also width right ?
If the pressure of the cell is fix and the quenching circuit works the way i proposed it becomes obvious that he stated 7.4ul per injection cycle because for every injection cycle you will inject the same amount of hydrogen. The amount of gas injected will be controlled by the rpm since as you add exhaust gas the RPM will change REDUCE automatically.
He probably had a sensor on the valves that every time the intake valve open the solenoid opens for 4ms injecting the gas as i said 13,3ml of hydrogen oxygen gas per injection...
than on the intake he probably used a kind of three way switch that has the air input and also the exhaust return... he than just regulated the speed rotating the switch... Isn't it very simple?
All is needed here now is to calculate the pressure needed to be maintained on the cell for having this amount of gas passing inside the quenching circuit in this predetermined time at a velocity greater than as he stated 350cm/second.
here is a link with a calculator i'm studying :
http://www.pipeflowcalculations.com/airflow/index.htmpressure in the cell is free so you can use it so no need for gas pump.
Right you also understood well the 2,5 times more powerful comparison is (by) volume but or also more probably or precisely (BY) weight...