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I have a cabin which is running off grid, using a solar panel and a (2*120 Ah 12V) battery.

I now want power in another building 20 meters away. One solution is to use an expensive high gauge (7 mm2)cable, to reduce the voltage drop - but that cable is also much more difficult to hide. I can't bury it without tripling the length.

I will have a bigger battery bank in the other building eventually, but the solar panels will remain on the main building.

When the voltage drops in the cable, is that also wasted Watt-hours? A few lamps will happily run on fewer volts. I am more worried about losing Wh's.

What other alternatives do I have?

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    \$\begingroup\$ Voltage drop will be proportional to current. What's the current? \$\endgroup\$
    – Transistor
    Jul 10 '16 at 10:30
  • \$\begingroup\$ Do you also want the battery in the second building to be able to send power back to the first building? \$\endgroup\$
    – Dave Tweed
    Jul 10 '16 at 11:13
  • \$\begingroup\$ There are commercial buck-boost converters for this very purpose. Simplest solution I can think of is 12->110/230 Vac converter, normal mains cable installation in between, 12 V batter charger and a battery and you are good to go. \$\endgroup\$
    – winny
    Jul 10 '16 at 11:27
  • \$\begingroup\$ @winny I assume that this will introduce a huge loss in efficiency. \$\endgroup\$
    – frodeborli
    Jul 10 '16 at 11:34
  • \$\begingroup\$ @DaveTweed Ideally, I would like to have all batteries in building 2, but I assume less current will go through the line with batteries in both ends. \$\endgroup\$
    – frodeborli
    Jul 10 '16 at 11:34
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The voltage drop is caused by the current. At 12 V you need a lot of current to transport any amount of power. P = V * I so for 120 Watt at 12 V the current will be 10 A. If the cables have a total resistance of 0.1 ohm (which is quite low for 20m) the cables will drop 10 A * 0.1 ohm = 1 Volt. So you lose 1 V * 10 A = 10 Watt in the cables.

A solution is to increase the voltage. Let's do that again but now use 120 V instead of 12 V:

120 Watt at 120 V gives 1 A, 1 A * 0.1 ohm = 0.1 V drop, 0.1 V drop * 1 A = 0.1 W lost ! So that's 10 / 0.1 = 100 times more efficient !

How to increase the voltage ? You could use step-up (boost converter) to increase the voltage and step-down (buck converter) to efficiently lower the voltage again to 12 V. On Ebay many of these modules are sold. It depends on the amount of current you need at 12 V which type will suit your needs.

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  • \$\begingroup\$ I am mostly curious of whether or not I am wasting energy. That is, will a light bulb in house 2 consume more watts per hour out of my batteries, than the same bulb in house 1? I assume it will be dimmer in house 2, but I am also hoping that it will consume fewer watts. \$\endgroup\$
    – frodeborli
    Jul 10 '16 at 11:31
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    \$\begingroup\$ Power is energy per second. So yes, you lose energy. The long cable will heat up a few degrees (nothing to worry about). The light in the remote house will burn less bright. The whole setup will consume less power in total (due to the cable resistance, less current flows) But you will also get less light in the remote cabin. It will consume fewer watts but you get less light per watt and that is wastefull. The simplest solution would be to use LED lighting in the cabins as these are much more efficient than bulbs. So you need less current and waste less power. \$\endgroup\$ Jul 10 '16 at 12:25
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    \$\begingroup\$ If using an up converter to go from 12V to the higher 120V as shown in your example it is really necessary to also consider the efficiency of the up converter. It may indeed be more efficient by 100 times just considering the wiring BUT an up converter is not ideal and even the best ones may not achieve 90% efficiency. So it becomes a tradeoff balance to determine if 12V distribution loss is greater or less than up converter distribution scheme taking converter efficiency into account. \$\endgroup\$ Jul 10 '16 at 14:37
  • \$\begingroup\$ For your scenario, you have about 91.7% efficiency when transporting power at 12 V. Boosting at one end and bucking at the other even if both converters are 95% efficient will have an overall efficiency of 90.2% as a fairly optimistic best case. So the converter option doesn't sound like a likely win. Of course the real answer depends on how much power is actually needed, and the actual efficiency of the available converters. OP needs to do some actual engineering rather than just guess what solution is most efficient. \$\endgroup\$
    – The Photon
    Jul 10 '16 at 14:53
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    \$\begingroup\$ The current should be fairly low, which means the resistance will be fairly low No, the resistance stays the same (it is a property of the wire) but the voltage drop and energy loss will be less. \$\endgroup\$ Jul 10 '16 at 15:28

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