OK, I think I see my malfunction. The calculator can't change the application itself in order to benefit from higher voltage, and it could only change % voltage drop? I remember that higher voltages can dramatically drop Amps and thus wire size to get the exact same power to the application. It always seems like it was getting something for nothing. You can't do that with water, for instance. You have either higher pressure or a larger pipe. Higher pressure means--at some point--thicker pipe, better fittings, etc.
Same thing; pressure is voltage, pipe is resistance, flow rate is current. Increase the pressure (same as increasing voltage) or pipe size (same as reducing resistance) and you get more flow.
Where the analogy breaks down for me is that you are getting more water pressure while using a thinner pipe, which doesn't work in hydraulics. The more pressure, the thicker the pipe. e.g., and I'm probably still not getting it, you can light a 120V 1000 watt bulb at 8.3 Amps. Using the same 10000 watt bulb with 240V, you only need to calculate for 4.1 Amps to supply the same power to the bulb.. So, you can use more pressure and a smaller pipe (wire) with electricity?
I believe you are confusing the "bursting" pressure of a physically thinner walled pipe ("thinner walled", not smaller I.D.) with pipe I.D.--if we define "smaller" pipe as just a smaller I.D. the analogy is spot on...
Using the 240V example above, you actually lower the current from 8.3 Amps to 4.1 Amps, but maintain the same power at the application. My mistake using the 240V example is that a 240 circuit is really two 120V circuits, which means the available current now has two feeds, not one. So you actually don't have smaller wire, just more of it, as in having a larger ID pipe with lower pressure but the same amount of water flowing through it. And that makes perfect sense to me.
Is that it?!
Home
Help
Search
Login
Register
