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Re: [PSUBS-MAILIST] electrical question
We had a brief discussion about the benefits of higher voltages last
January. A relevant discussion (using an example of a 150 amp current
at 36 volts) is copied below:
The resistivity (rho) of any material is a temperature dependent
variable, but the variance is small, and values for many materials at 20
degrees C are available on the internet. A quick search provides the
resistivity value for copper as 1.724E-8 ohm*meters.
Resistance, resistivity, length and cross-sectional area are related by
the equation
R = (rho * l )/A
Voltage, current and resistance are related by the equation
V=I*R
Combining the two, and rearranging for A gives
A = (I * rho * l ) / V
Now, it is important to note that the V above is not your nominal
operating voltage, but rather a voltage corresponding to the acceptable
voltage drop in the wire. Many electrical codes give this as around 2%
of the nominal voltage, so for a 36V system that would be 0.72 V. You
didn't provide a wire length, but for the purpose of discussion, for a
10 meter wire, you have
A = (150 * 1.724E-8 * 10) / 0.72 = 3.6E-5 m^2, or 36 mm^2.
The corresponding AWG wire gauge is about 2 AWG.
This is large, but this would be necessary if you were running at full
power continuously, and could not tolerate voltage drops higher than
0.72V. Provided you stay within safety limits (i.e. the isolation
voltage limit of the wire insulation), and can tolerate more heat, and
more voltage drop in the wire at maximum current, you can derate this to
be a more appropriate fit to the expected continuous load.
Now, with a 150 amp current delivered at 36 volts, your VA is 150 * 36 =
5400. Now, let's say you were to deliver that same power at double the
voltage, or 72 volts. For the same VA, your current would then be
halved, or 75 amps. What happens at the elevated voltage is that the
acceptable voltage drop increases, since 2% of 72 is greater than 2% of
36. Performing the required area calculation for the same 10 meter wire
in this case gives:
A = (75 * 1.724E-8 * 10) / 1.44 = 8.979E-6 m^2, or 8.979 mm^2.
Note that although we only doubled the system voltage, we reduced the
required wire area by a factor of 4. The corresponding AWG wire gauge
is about 8 AWG.
This is obviously much more economical in terms of wire, switch and
conductor cost, so it should come as no surprise that long-distance
overland power transmission lines use potentials in the hundreds of
kiloVolts, stepping down as appropriate using transformers (substations)
close to power consumers.
-Sean
vbra676539@aol.com wrote:
No problem. The 12 volt motors will produce less thrust at maximum
draw than the larger units, but otherwise it's not a problem. The
issue is efficiency. Your wiring must be larger for equivelant draws
at lower voltages. Heat losses, resistance in the system and
overworked little motors will probably cause issues down the line.
Study up on DC electrics to see how all that works, and how it
improves things to go up in voltage.
Vance
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