I am trying to send power from a system to a load that is almost 200 meters aways. Now, the supply from the battery is approx 70V or so. At the load end the voltage received is 55V. Now, I understand that the power cable will have its own losses due to which there will be a drop.
But I want to understand the mathematics of this drop and the equations involved in it. Can anyone direct me to an app note for this please ?
As others have explained, it is a simple calculation using Ohm's Law. Two methods of transmitting power over a long distance include...
Remote Sense. Where the power supply uses an additional wire to sense the voltage at the far end (where the load is). Then the power supply adjusts its output to compensate for the losses through the long distance.
Remote voltage regulator (or converter). Where a higher voltage is sent through the line and a regulator is added at the distant end where the load is. This has the advantage over remote sensing because it doesn't require another wire way out to the end.
The mathematics is Ohm's Law, V=IR.
It's exactly proportional, voltage drop proportional to current, and to resistivity, which is inversely proportional to copper cross section.
I use a rule of thumb for the resitivity of copper at room temperature, which is 17mohm per meter of 1mm2 cable. Copper increases resistance quite markedly as it warms, +10% for a 25C temperature rise, so that 17mohm is given to all the accuracy needed.
Let's say you were transporting 1 amp to your load through 200m of 1mm2 cable, and back of course.
400m of 1mm2 cable, by my rule of thumb, will have a resistance of 400*17m = 6.8 ohms. It will have a drop of 6.8 volts at one amp, 13.6 at two amps, over 20v at 3 amps.
Ordinary house wiring 1mm2 cable is actually rated for 10A. However, that's a heating spec, not a voltage drop spec. It's expected that runs will be short, and at 240v, a bit of voltage drop is small compared to the supply voltage.
If you use (say) 4mm2 wire instead, your resistance and voltage drop will be one quarter of previously, but your cable cost will probably be not quite 4 times.
The easy solutions for low drop are therefore ...
a) to use less current (which may not be possible)
b) to use more copper cross section (which will cost money)
The more complicated solutions for low drop are ...
c) step the voltage up before you send it, and step it down at the far end (the economics of this get better when you have a very long distance, just ask the power companies, when the cost of the conversion (money and electrical efficiency) is less than the cost of a thicker cable)
d) step the voltage up at the receiving end so it's right
e) step the voltage up at the sending end, so it's right at the far end (slightly more efficient than (d) due to the lower current, but you'll need remote sensing)
