Using clothesline steel core wire rope for AC and DC

Question

As I am currently in a war zone, I don't have many options for cabling.

I found this clothesline (steel core plastic wire rope) that appears to be one mm of diameter (steel core diameter.) 13 meters of it measured 7 ohms resistance.
Edit: It is 3.8 Ω and not 7. The first multimeter test lead probes had 2-4 resistance when shorted. A slightly better multimeter had 0.5 Ω when shorted. Both multimeters gave 3.8 after subtracting multimeters own resistances and scratching the wire ends.

1. Can it carry AC 120 or 240 volts, and if so, for what distance?
2. How many of it (doubling it) can carry DC 18V and 15 A from a solar panel arrays 10 to 15 meters away from the inverter (charge controller)? (20 W panels with open circuit voltage of 21 V).

This is just temporary solution and I hope only for few days or weeks. Air strikes blew up some transformers and high voltage lines and our concrete homes are not designed to be habitable without AC power.

Edit:
Its now connected to a c32 breaker (the smallest I could find) and a breaker mounting brackets cut from a laptop battery cover. The wire is inserted in plastic bottle caps as wire wall clips and a tow heads plug is inserted to the other end (to be upgraded to three heads) because now polarity is important I think.

I will connect it to a manual changeover switch when adding the solar oart after finishing the battery but that is another longer story (Lithium battery without BMS) but I may be able to reuse old laptop batteries BMS (I have more than 20 batteries):

The solar system is only for a medium 100 W Samsung fridge and the mains are for the fridge plus two ceiling fan, one swamp cooler and three LED lights.

Answer 1

Steel, having just around 10 times higher resistivity than copper means it will take ten times the conductor area to match copper. If you measured 13 meter of it to 3.8 Ω, the cross sectional area would be 2 mm^2, assuming 5.95*10^-7 Ωm of resistivity for "high alloy steel" (this varies greatly unfortunately so assume +100% -50% uncertainty for all values given).

To answer your questions:

1. How long = Time: probably many years. How long = Distance: Most devices will run happily with 10 % voltage drop. Anything universal input (100-240 V) could handle significant voltage drop due to the cable at which point it's the thermal capability of the cable which sets the limit as you don't want it to melt.

   With 3.8 Ω for 13 meter, you have 0.29 Ω/m. At 1 A 230 V AC current, you can go 39.7 meter (round trip is double distance) before you have dropped 10 % of the voltage. If you halve the current, it's double the distance. Gut feeling + experience says it would get lukewarm at 2-3 A so I would not go much above it.

Could be lethal though, as the insulation is not mains voltage rated. I would be more afraid of anyone coming into contact with the end points and the cable termination than touching the outer shell of the clothesline and somehow get zapped by it as they tend to sit outside for decades without becoming brittle by the UV exposure.

As stated below by Martin McCormick, your best bet is to put the inverter as close to the panels as you can and run AC through the clothesline versus low voltage DC current through the clothesline.

2. With just one conductor, carrying 15 A via 3.8 Ω means a 57 V drop, so not possible with 18 V at all. It would also melt. To make it work at 18 V, perhaps 20 % drop (3.6 V) could be tolerated. To get down to 3.6 V drop, you would need 57/3.6 = 16 in parallel.

Answer 2

Agree with winny to put the inverter next to the solar panels if possible and send 230V AC (standard in Sudan) over the clothesline, despite it being more dangerous.

If you need to transfer 1000W of power, it loses less to clothesline resistance when transferred at a higher voltage.

1000W =

• 55 amps @ 18 volts

• 4 amps @ 230 volts

Power lost through the clothesline is (Current)^2 * (Resistance).

So choosing the lower 4 amps current is much less loss (~200x). The power loss dissipates as heat in the clothesline, at some point, it will burn, so it's not just about efficiency--using 230V is necessary to get this to work. You can probably carry 1-2kW with 1mm core wire at 230V AC. This is a window-unit air conditioner and some light bulbs or a couple of large fans. Maybe a microwave for short periods of time. Carefully observe for smoke of the clothesline insulator and shut it off immediately. If this happens you need to run less equipment in the house at the same time or run more parallel lines.

Use your home-harvested copper wire for the 18V DC connection between the panels and the AC inverter.

Answer 3

1. The AC side should be OK, as others have noted.

2. As winny notes, you'll have a problem trying to carry 15 DC amps. However, I see you're using an array of 20W panels. You said 18V and 15A, so 270 watts; I'm going to round and guess you have 14 panels in parallel. Can you connect them in series instead of parallel, giving you a high voltage and low current?

• 14x1 configuration would give you 252V and about one amp, a 7V drop over the measured 7 ohms or 3% loss.
• 7x2 would be ~125V at 2A, a 14V drop so still only about 11% loss.

You'll of course have to check whether your transformer can handle the input. You'll also have less shade tolerance than in the parallel configuration, but that sounds like the least of your worries.

I just used your measurement at 13 meters for resistance. 7/13 is about .54 ohms per meter. If the panels are 15 meters away from the inverter, that's a 30m round trip, so closer to 16 ohms. That's still probably fine, or one doubling gets you back to the numbers above.

Answer 4

I'm not an electrician, but I can help with the circuit analysis. The formula for resistance is:

$$R = \rho \frac L A$$

where \$\rho\$ is the resistivity of the steel, \$L\$ is the length of the wire, and \$A\$ is the cross-sectional area of the wire. The first question is whether your resistance measurement is correct. Measuring small resistances with a two-wire multimeter is hard due to the resistance of the leads. You also (for testing and use) need to make sure you have good contact with the wire. The surface of metals oxidizes, which can add resistance to a weak connection.

Your measurement of 7 ohms for 13 meters give a resistivity of:

$$\rho = \frac {R A}{L} = \frac {7\Omega \cdot \pi \cdot (0.5 \mathrm{mm})^2} {13 \mathrm m} = 4.23 \times 10^{-7} \Omega \cdot \mathrm m$$

which seems to be in the right range according to this and other sources, so your measurement seems reasonable. But you didn't say how you're doing the measurement.

Resistance is linear with distance -- halve the length of the wire and you halve its resistance. Double the length of the wire and you double its resistance. Two wires in parallel will have half the resistance of one, three wires will have a third of the resistance of one, and so forth. Your wire seems to be \$0.5385\ \Omega/\mathrm m\$. If you want to make a cable with around 1 ohm of resistance, you could use:

• 1.85 meters of 1 wire
• 3.7 meters of 2 wires in parallel
• 7.4 meters of 4 wires in parallel
• etc.

If you want to carry your large 15 amp current, you'll need to go lower. Maybe a 0.1 ohm cable can work:

• 1 meter of 6 wires in parallel
• 2 meters of 12 wires in parallel
• 4 meters of 24 wires in parallel
• etc.

For comparison, copper's resistivity is around \$1.7 \times 10^{-8} \Omega\cdot\mathrm m\$. Your steel's resistivity is around 25 times higher.

Aluminum is almost as good as copper at \$2.7 \times 10^{-8} \Omega\cdot\mathrm m\$, so if you find any aluminum wire (or long strips of aluminum if you get desperate) that would be much better than steel. Be sure to insulate -- high currents can be just as dangerous as high voltages if they get shorted out or the connection suddenly breaks.

Good luck! I hope your situation improves soon.

Answer 5

Source: https://news.ycombinator.com/item?id=35730074

Credit: https://news.ycombinator.com/user?id=ltbarcly3

TLDR: connections must be protected by some kind of anti oxidation coating, if you have nothing else use grease but something designed for electrical connections is better. If you have nothing else, melt some lead and dip the exposed part of the wire in that to coat it. Lead should be readily available in a war zone? Long term the wire WILL melt at some random point along the wire so it is much better if this wire is kept away from anything flammable.

The risks of using steel cable for power are going to be corrosion/rust. The primary way this will cause problems is around connections, but long term it is also an issue away from connections. At connections, corrosion can cause the wire to become loose, something like ox-gard can be used to delay this significantly. It isn't designed for steel but some kind of protection must be done at connections or arcing will be a major risk. Longer term the wire itself will rust, and at some point the conductive cross section of the wire will be compromised to the point it overheats and melts. This might be a year and it might be 100 years depending on factors which are hard to predict and control.

Answer 6

If you have access to lead or Tin.

1. Strip cloth line to get the steel core (Blade) (If burned, scrub so the outer oxidation is removed).
2. Vasiline or try other available items to be used as a flux. Cover steel wire with the hack-flux.
3. Run Dip the wire in lead , you can also pull the wire through the molten lead.
4. Once lead is cooled, do the same with enamel or use carefully keeping distance between +/-

The outer lead will do the conduction, inner steel provides the mechanical support.

Answer 7

In a war situation you use everything that is available. You should just be cautionous for:

• voltage drop because of the high resistance
• bad contacts because of oxidation
• overheating because of the high resistance
• failure of the insulation because of overheating
• dangerous situations because the plastic is not really designed as isolation and may not be able to withstand the actual voltages