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I've lost my 'intuitive feel' for current, voltage and power in a conductor (like a 18 gauge copper wire)...

The wire seems to be rated by maximum current rather than maximum power... Does that mean I can get a 'free ride' by using a high voltage in order keep the wire size low when transmitting power?

Simple example: let's say Im going to run a 18 gauge wire around my house to supply power to a bunch of 12 V devices... Can I actually support a maximum of 4x more devices if I supply 48 V in the wire system and then step it down at each destination.

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    \$\begingroup\$ "does that mean i can get a 'free ride' by using high voltage in order keep wire size low when transmitting power". Yes, that's why the power utilities step their long transmission lines up to 10's or 100's of kV. \$\endgroup\$ – brhans Mar 20 '15 at 16:47
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    \$\begingroup\$ ... not QUITE a free ride, but enough savings to more than pay for the transformers etc at each end. Whether you get enough savings between 12V and 48V you'll have to work out for yourself but your thinking is correct. \$\endgroup\$ – Brian Drummond Mar 20 '15 at 17:01
  • \$\begingroup\$ Power losses for AC due to capacitance can be significant at high voltage, especially for hundreds of kV submarine cables. \$\endgroup\$ – Spehro Pefhany Mar 20 '15 at 17:30
  • \$\begingroup\$ Another thing you need to consider is losses that occur in the converters. Doing it right can make all the difference between getting a return on investment or just pouring money down the drain. \$\endgroup\$ – Sean Boddy Mar 20 '15 at 20:33
  • \$\begingroup\$ PoE technology is basically this - it starts up at around 48v and steps it down to ~12v or 5v depending on what the device at the other end needs - the caveat is that the step-down is usually a DC-DC converter and works just as well if it receives 48v vs 20v - it will accommodate relatively large losses in the wire. \$\endgroup\$ – user2813274 Mar 21 '15 at 1:32
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As a general rule of thumb, the current determines the thickness of the wire, and the voltage determines the thickness (and/or material) of the insulation.

The power grid does pretty much what you propose - use much higher voltages to reduce cable size. The reduced current also means reduced line losses, which is very important over long distances.

For instance, if you had 1kW of power you you wanted to move from A to B, you could use, say 100A at 10V, or maybe 10A at 100V.

For 100A power transmission you'd need 1 AWG wire. That has a diameter of 7.34822mm, and a resistance of 0.406392Ω per km. So over a 1km distance you'd lose 0.406392 * 100 = 40.64V. Ouch. That would just plain not work! So although the cable could physically cope with that current, over that distance you'd loose all your voltage. So that would be a no-go.

Try at 100V, 10A.

10A can go through 11AWG cable. That's 2.30378mm thick, and with a resistance of 4.1328Ω/km. Much higher resistance, but much lighter cable. How much voltage would we lose over 1km? 41.328V. Factor in the return path, so you double the distance, you end up losing 82.656V, leaving 17.344V left for the load. Getting there. Still not workable, but getting there. That equates to 173.44W.

How about if we pump it right up to 1000V, at just 1A? At 1A we can use 21AWG wire, at 0.7239mm thick. 41.984Ω/km, which would be 41.984V lost there, and 41.984V lost back. So 83.968V lost from your 1000, leaving 916.032V. That's 916.832W coming out.

Now supposing you wanted to transmit 100A over a 1km distance, and limit the voltage drop to say no more than 1V. What thickness of cable would you need for that? Well, for a 1V drop at 100A you would have a resistance of 1/100 = 0.01Ω. So your cable must have no more than 0.01Ω/km. The table I use doesn't go that low, so we'd need to do some calculations.

If we use copper wire, that has a resistivity (\$\rho\$) of \$1.68×10^{−8}\Omega/m\$ at 20°C. For resistivity we have the formula: $$ R=\frac{\rho L}{A} $$ where L is the length (1km), R is the resistance (0.01Ω) and A is the cross-sectional area.

So we can re-arrange that for A: $$ A=\frac{\rho L}{R} $$ and substitute our values: $$ A=\frac{1.68 \times 10^{−8} \times 1000}{0.01} $$ $$ A = 0.00168m^2 $$ And of course, that equates to a wire diameter of 4.6cm.

Is a 5cm thick cable practical for that? Not if you can increase the voltage to decrease the current, no.

So for power transmission it is possible to reduce the cable size over very long distances and reduce the losses over the line.

Over shorter distances the losses are considerably less, but can still be a problem at higher currents. But is it worth it, or better to just use fatter cable?

You also have to factor in:

  • The efficiency of the power conversion - step up / step down.
  • The cost of better insulation if you have very high voltages.
  • The reduction in cost by using thinner cable.
  • Safety issues - higher voltages are dangerous.

So is it a "free" ride? No. There will always be losses and caveats you have to look out for. It can, though, get around problems of longer distance transmission of power.

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  • \$\begingroup\$ Instead of calculating resistance with a given wire thickness, why not calculate the wire thickness you'd need for a given efficiency (say, 10% loss)? \$\endgroup\$ – user253751 Mar 21 '15 at 2:03
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    \$\begingroup\$ You said 10kW initially but all your examples are 1kW. \$\endgroup\$ – R.. GitHub STOP HELPING ICE Mar 21 '15 at 2:09
  • \$\begingroup\$ I was in a hurry. Dinner was ready. I have changed it to 1kW \$\endgroup\$ – Majenko Mar 21 '15 at 10:50
  • \$\begingroup\$ @immibis Good plan. I have added a nice spot of calculation for a really fat cable. \$\endgroup\$ – Majenko Mar 21 '15 at 11:45
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wire seems to be rated by max current more than max power... does that mean i can get

a 'free ride' by using high voltage in order keep wire size low when transmitting power?

In the sense you're talking about, yes. And that, as a matter of fact, is why Edison was wrong, Westinghouse was right, and we use AC rather than DC.

The limit on wire size is self-heating. For a given length of wire which is supplying current to a load, the amount of current flowing through the wire can be independent of the power being used by the load. That is, if you've got a 12-volt car battery driving 10 amps, and a 220 volt circuit driving 10 amps, the first supplies 120 watts of total power and the second supplies 2200 watts, but 10 feet of 10 ga wire (total resistance 0.01 ohms) will dissipate 1 watt (I squared R = 10 x 10 x 0.01 = 1) regardless of the total voltage.

And that, of course, is why long distance power transmission always uses high voltage - the higher the voltage the lower the proportion of power lost to the transmission line for a given wire size.

As Majenko points out, this isn't exactly a free ride, in that the wire will always dissipate some power, but for a given wire size and length you can always provide more load power by increasing the voltage (as long as your insulation holds out, of course, but that's a different story).

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  • \$\begingroup\$ great explanations but that leads me to this Q.... how does the load at the end of those 10ga wire's "feel" the voltage being applied? i.e. when i send 10A of current down a wire with 24V vs 10A with 12V i am able to power a load of 2x the power... but its not like the current flows twice as fast! current is, by definition, already the flow rate of charge. \$\endgroup\$ – Cool Pontiac Mar 21 '15 at 18:12
  • \$\begingroup\$ @CoolPontiac - Yes, but power is the product of voltage and current, so for the same current if you double the voltage you double the power. And you don't "send 10 A of current down the wire" in the way you seem to think. The total voltage, divided by the sum of all the resistances in series (wire, load, return wire) is what determines the current level. For the same total resistance, if you double the voltage, you double the current. \$\endgroup\$ – WhatRoughBeast Mar 21 '15 at 23:16
  • \$\begingroup\$ thanks to all for great answers. i see now that using higher voltage to deliver higher power at end of a fixed size wire is not getting a free ride ie. free 'work' because you have to design a load that provides higher resistance in order to generate more 'work' from that max current that your fixed wire can carry. it seems the "water analogy" fits here. you have a given size pipe pushing water against a filter at the end. you make the filter much thicker so that the same flow of water does more 'work' but you now need to put more pressure on the water to make it flow at same rate. \$\endgroup\$ – Cool Pontiac Mar 23 '15 at 16:22
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FYI: Transformers only work on AC not DC. If you are running a DC line like in a car, you have to have thick wire or use DC to AC and AC to DC converters

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