TomGeorge:
"So you are designing a household standby power system, for how long does your mains power drop out?"
A few days at a time, maximum, usually in severe snow storms or high winds, etc.
However, considering the possibility of an EMP or other type of terrorist attack, where the grid could be down indefinitely, I want the unit to be able to work under those circumstances as well.
Needless to say, the device's electronics and the alternator will be housed in a good Faraday grounded cage. I am aware that some types of household appliances will be destroyed by an EMP attack, if it ever occurs, but resistive heat, hot water heater, incandescent/halogen lights, and hardened appliances, etc. will not.
This device is a "looped supply". By that I mean, the water powered motor (WPM) is a jetted rotating drum, that rotates inside an enclosed sump box and with a rotating shaft drives an external alternator via pulley and belt. This is NOT a free-energy situation, just using old technology in a slightly different way.
The drum has water, from the bottom of the sump box, pumped into it. The jets along the outer edge of the drum cause the drum to rotate. Water pumped into the drum by an electric water pump, which in my case pumps out 40 gallons/minute (16 ft/s) to the drum.
So along with the proposed proportional valve, to control the drums speed below maximum safe speed, there will be an on-off manually adjustable valve that is set to limit/restrict the maximum amount of water flow to the drum, when the alternator has no load or if the pulley belt connecting the WPM and the alternator breaks, etc.
Robin2:
The reason I am not considering running my household on an inverter system, powered by lead-acid batteries, is because of the cost of the batteries, their potential danger, and the frequency of needed battery replacements and the number of needed batteries, if they are going to be treated properly to prevent early battery destruction.
By that I mean, a lead-acid battery should not be discharged at a rate (I am told) faster than 1/20th of its AH rating.
That means, if I were to use 1000 AH deep cycle batteries and observed the 1/20 rule, each battery should only supply a 50 AH rate of discharge to the inverter.
So, to be able to power the "full" 7,200 Watts of inverter power, I would need to have 12 batteries in parallel.
Now, I don't know about you, but for me THAT is NOT a quantity of batteries I can easily afford to purchase on my very limited Social Security income.
With batteries of a lower AH rating it would require even more batteries to supply the full maximum power for the inverter at even greater cost.
PerryBeddington:
"You ask if the output current is controlled by the speed of the alternator shaft. Two answers and one question:
The question is, why do you think this is the case?"
Because I was told, "The amount of power it takes to turn the alternator increases as the current load increases as the battery is being recharged or electronic devices in the vehicle require them. The number of RPMs required to produce 50 watts or 100 watts could be 2000 RPMs yet the engine will be working harder, burning more fuel, to generate 100 watts."
So, to me, that means, that as the load current increases, then the drag on the alternator's shaft increases, causing the motor to have to put out more power (torque) to maintain the same needed 3600 RPM speed of the alternator's shaft.
I am not trying to increase or decrease the speed of the alternator's shaft to "control" the output current, but rather to maintain the 3600 RPM to keep the alternator putting out 120 VAC @ 60 Hz for up to 7200 continuous watts.
I think that as the drag on the alternator's shaft varies as the load requirements vary, then that varying drag will put a corresponding varying drag on the motor's shaft which would vary that motor's speed, which in turn would vary the alternator's speed. Right?
Thus I need to be able to regulate the motor's speed by altering its water input speed to maintain it at the constant 3600 RPM speed needed for the alternator. Right?
"Do you know Ohm's law?"
I am well aware of Ohm's law.
" The load takes whatever current it needs when you apply the appropriate voltage. The alternator does not control the current, the load does."
The alternator does determine the maximum current it is able to put out at 120 VAC @ 60 Hz. Right?
Now for lesser amounts of needed current, then the load draws what it needs. But as it does so, it varies the drag on the alternator's shaft, which in turn, varies the drag on the motor's shaft Right?
Paul__B: Please see answer for Robin2 above.
As I said earlier, I have no prior experience with portable power supply alternator heads, so PLEASE clear up any and all misunderstandings I have as stated in my answers.
Rick