RE: [CAD_CAM_EDM_DRO] Re: welding machine as a power supply [Off topic enough yet?]

>Roy: Draw it out on paper. [...]
>
>This has nothing to do with transmission line theory. In that case you

are starting with a fixed

>resistance per foot. At a given load (in Watts) the voltage drop

across the series resistance will remain

>the same regardless of the voltage. So at 120V the voltage drop across

twenty miles of line might be 50V.

>At 10,000 volts the same load (in watts) still has 50V which is a much

smaller percentage of the total.

I think it's time for everyone to tune up their slide rules...

For a given resistance (e.g. a power transmission line),
If at 120V (at the load) the transmission line drop under a given load
might be 50V,
then at 10kV (= 83.333*120) (at the load) the same load (in watts) will
need 83.333 times less current,
and the same resistance will only produce a 0.6 volt drop (not 50 as
mentioned above) and the power consumed by the line
will be reduced by a factor of 83.333^2 = 6944

For a simpler example, if your line resistance is 1 ohm, and you need
100 Watts at the load,
and you can rewire the load (using a transformer, resplicing motor
windings, etc) to run on either
100 volts or 1000 volts, you get the following two cases:

Load supply voltage: 100V 1000V
Current: 1000W/100V = 10A 1000W/1000V = 1A
Load resistance: 100V/10A = 10ohm 1000V/1A = 1000ohm
Line voltage drop: 10A*1ohm = 10V 1A*1ohm = 1V
Line power loss: 10V*10A = 100W 1V*1A = 1W
Line supply voltage: 100V+10V = 110V 1000V+1V = 1001V
%Line power loss: 100W/(100W+1000W) -> 9.1% 1W/(1W+1000W) -> 0.1%
Notice that the power loss has been reduced by a factor of 100.

Didn't this thread go Off Topic a long time ago???