cmhollis said:
Nexdog,
Are you familiar with Ohm's law? Power (Watts) = Volts (V) x Current (I).
So, if your pump is rated to draw 300 watts, at a lower voltage (100 vs 120), it is going to draw more current to maintain the power necessary.
Not exactly.....
The wattage is given by Watt's Law P = I * V, but you can't assume that it uses the same power (wattage) for any given voltage. You would first need to apply Ohm's law (V = I * R ) to determine the current drain for a given input voltage, then plug that number in to determine the wattage.
Assuming a pump is rated 300W at 120V, that's 2.5 Amps (using Watt's Law). If we plug those numbers back into Ohm's law, we can estimate the pump motor's equivalent resistance as 120/2.5, or ~48 Ohms. It's a bit more complex in reality (EE pun there), as electric motors actually present inductive loads, but this is good enough to use as a equivilent resistive load for these calculations (Thevenin's theorem).
So, assuming 100V power in Japan, how much current will the pump draw?
100V = I * 48Ohms
I = 2.08 Amps
Then the wattage would be
P = 100V * 2.08A
P = 208 watts
I'm sure everyone will agree that water flow will be SIGNIFICANTLY reduced running at 208 watts vs. 300 watts on US 120V.
This is a nasty problem to have to deal with NexDog, you have my sympathies. As someone else suggested (was it melev?), I would look into a large transformer to provide 120V power to the tank circuits, and even consider buying a 100V backup pump in case you have problems with the transformer circuit. That way, you can at least keep the tank running. Redundancy, redundancy, redundancy.....