How to build a lightweight 50A extension cord

[FlasherZ]

We only count half since it's the same circuit... and #8 has 8.34mm cross section. Wish I did this before I bought the cable.. Could have saved another 4-5lbs probably.

Hmm? No, you don't count half, because the table is rating the entire cable with 2 cores for supply and not an individual conductor.

The IEE tables say you could use #10, but that's stretching things a bit far, especially with continuous load. #8 should be fine without issues (roughly 80A according to IEE).

 

[DMC-Orangeville]

Bringing this back from the dead, there is 6/3 SOOW "non-UL" cable available that 80% of the weight the one OP got. As far as I can tell the UL listing is just for physical abuse and flame resistance? Should I be worried about something else? Given it's for emergency use, I think it should be fine. Although buying it from ebay, it's probably not really UL listed anyway.

There are 2 possibilities as to why it's not UL listed
1. The cable may be old. It was common a number of years ago, to sell a cable for non-continuous load (eg arc welding). It allowed cable manufacturers to use less material, as the heat generated would be on a duty-cycle only (on-off-on-off etc). this would allow the cable to cool down. That explains the lighter weight. This cable went out of vogue 15-20 years ago...and I have worked for many cable manufacturers in 35 years (Carol, ITT, Royal, Triangle, General, American Insulated Wire, and Priority), as most electrical inspectors demand a UL listed (or CSA in Canada) on any jobsite or industrial workplace.

2. The cable has never been submitted to UL for testing, and may not pass, for many reasons. This could also explain the lighter weight. It may not even be 6 AWG.

In either case, I'd take a pass on this, and just go to a local electrical wholesaler, and buy a cut length of 6/3, 8/3, 6/4, 8/4 SOOW cable. It's usually available from stock

 

[AWDtsla]

Hmm? No, you don't count half, because the table is rating the entire cable with 2 cores for supply and not an individual conductor.

The IEE tables say you could use #10, but that's stretching things a bit far, especially with continuous load. #8 should be fine without issues (roughly 80A according to IEE).

They are simplifying heat sinking. But both pairs have equal current and therefore equal heat output, so it's not really necessary to think about both directions. Free air is kind of meaningless, since no one rates the heat sinking capabilities of the cable jackets, just the temperature where they melt. You could get twice the current through a cable that has thermally conductive jackets versus some other cable.

 

[FlasherZ]

They are simplifying heat sinking. But both pairs have equal current and therefore equal heat output, so it's not really necessary to think about both directions. Free air is kind of meaningless, since no one rates the heat sinking capabilities of the cable jackets, just the temperature where they melt. You could get twice the current through a cable that has thermally conductive jackets versus some other cable.

I'm still not understanding the point you're trying to make with halving the current.

The table provides the current rating for the entire cable under a single-phase power supply scenario. It assumes the current is equivalent on both cores - it's not a rating for each conductor.

So in the case of Tesla's use of 2 each 2.5 mm^2 conductors for each leg, it's the equivalent of a 5.0 mm^2 conductor for a single leg. The table's column 1 represents the cross-section for a single core, so you look in column 1 for the best approximation of 5.0 mm^2, and the rating (for whichever reference model you use) corresponds to the entire circuit current (as you say, don't think about both directions).

If you use #8, you do the same thing, but look for 8.37 mm^2 in column 1 and you'll approximate ~80A or so on a #8, two-core, single-phase cable (although the rating table for rubber cord is less than some of the other insulation types used @ 60ish).

What am I missing in regards to halving the current?

 

[AWDtsla]

I'm still not understanding the point you're trying to make with halving the current.

The table provides the current rating for the entire cable under a single-phase power supply scenario. It assumes the current is equivalent on both cores - it's not a rating for each conductor.

So in the case of Tesla's use of 2 each 2.5 mm^2 conductors for each leg, it's the equivalent of a 5.0 mm^2 conductor for a single leg. The table's column 1 represents the cross-section for a single core, so you look in column 1 for the best approximation of 5.0 mm^2, and the rating (for whichever reference model you use) corresponds to the entire circuit current (as you say, don't think about both directions).

If you use #8, you do the same thing, but look for 8.37 mm^2 in column 1 and you'll approximate ~80A or so on a #8, two-core, single-phase cable (although the rating table for rubber cord is less than some of the other insulation types used @ 60ish).

What am I missing in regards to halving the current?

Not halving the current. The current is the same in each wire of a 2 conductor cable, and each conductor puts out half the heatload over the length of the cable. Those tables are just an approximation of the thermal dissipation limit of that configuration cable, of course with a large safety margin built in. You can do a simple calculation, take the voltage drop over the length of the cable, multiple by current, and then double for the conductors, and you get how much heat has to be dissipated along the length of the wire to keep the temperature change small, because P=IV.. or use expected resistance to calculate the same. If you do something like Tesla did, and liquid cool your supercharging cables, then all those tables become completely meaningless.

Since we'll never find the right data from manufacturers per cable, the best thing to do is just to do it empirically. Get a particular cable, run it at x amps until the temperature reaches steady state, check if that temperature is safe. Anyways, clearly 6/3 is way overkill. If I get bored I might order some different types of cable...

 

[Cottonwood]

Actually, it seems we're both wrong. And my bad recollection comes from further up in that thread, where I guessed at 2x #14.

Tesla Model S UMC cut open and modified to J1772 - Page 2

Ingineer says they're 2.5 mm^2, which is roughly #13.

So, back to my trusty IEE table 4E4A, a thermoset cord utilizing a 5.0 mm^2 conductor for each leg would be rated at roughly 59A or so in free air.

Each leg uses 2 power conductors, ground uses 1 power conductor, and the 2 18ga are used for the signaling (proximity / pilot).

Another good approximation is that 2 parallel conductors are the equivalent of a single conductor with a wire gauge 3 numbers less* (larger wire). For example, 2 parallel #13 wires is the equivalent of a single #10 wire.

I have done testing using a #10 extension cord at 40 Amps with a UMC. My ad hoc test showed that a #10 extension cord has acceptable temperature rise if the cord is laid out across the floor (free air). The problem happens if you leave the cord coiled up. With a coiled cord, I found the temperature rise unacceptable (sorry, I did not record quantitative results). This makes sense because there is less total area to dissipate the total heat of the multiple passes of the cord laying next to each other.

If you do use a smaller size extension cord, or even the UMC, avoid coiling the cable too many times, or too tightly.


* The AWG scale is logarithmic similar to dB. A change of 10 wire gauges is a change of a factor of 10 in wire cross section. A change of 3 is 10[SUP](3/10)[/SUP] or 1.995, approximately 2.

 

[FlasherZ]

Not halving the current. The current is the same in each wire of a 2 conductor cable, and each conductor puts out half the heatload over the length of the cable. Those tables are just an approximation of the thermal dissipation limit of that configuration cable, of course with a large safety margin built in. You can do a simple calculation, take the voltage drop over the length of the cable, multiple by current, and then double for the conductors, and you get how much heat has to be dissipated along the length of the wire to keep the temperature change small, because P=IV.. or use expected resistance to calculate the same. If you do something like Tesla did, and liquid cool your supercharging cables, then all those tables become completely meaningless.

Since we'll never find the right data from manufacturers per cable, the best thing to do is just to do it empirically. Get a particular cable, run it at x amps until the temperature reaches steady state, check if that temperature is safe. Anyways, clearly 6/3 is way overkill. If I get bored I might order some different types of cable...

With the tables, I don't understand why you have to calculate heat dissipation and deal with all the complexities of insulation-trapping, etc.

My point is that the IEE tables give you a current rating for a particular type of cord that will maintain integrity of its insulation. It's based on the cross-section of one of the line conductors but provides you with a current rating for the entire cable. So no need to calculate any heat load or dissipation or anything like that - simply use the tables to identify the minimum cross-section size of one power conductor for a required current, or identify the maximum rated current for a given size of power conductor.

 

[Oba]

Actually, it seems we're both wrong. And my bad recollection comes from further up in that thread, where I guessed at 2x #14.

Tesla Model S UMC cut open and modified to J1772 - Page 2

Ingineer says they're 2.5 mm^2, which is roughly #13.

So, back to my trusty IEE table 4E4A, a thermoset cord utilizing a 5.0 mm^2 conductor for each leg would be rated at roughly 59A or so in free air.

Each leg uses 2 power conductors, ground uses 1 power conductor, and the 2 18ga are used for the signaling (proximity / pilot).

Can you provide a source that 2.5mm is 13 gauge?

 

[AWDtsla]

Can you provide a source that 2.5mm is 13 gauge?

Diameter vs area. i.e. Pi*r^2

 

[Cottonwood]

Diameter vs area. i.e. Pi*r^2

Can you provide a source that 2.5mm is 13 gauge?

Pi*(1.8288mm/2)[SUP]2[/SUP] = 2.6268 mm[SUP]2[/SUP]

That is approximately 2.5 mm[SUP]2[/SUP].

 

[ManjuM]

I bought a CAMCO 30' 14-50 Male to Female extension cord on Amazon for $106.44 or so, I was not happy with it for a number of reasons, here is a link to the cord being discussed: Camco RV 55195 50 Amp 30 Extension Cord with Handle : Amazon.com : Automotive

It's way too heavy: since it's for a 14-50, it has an "extra" #6 AWG conductor in it (for the neutral), which is not needed for EV charging. Adds weight and bulk.

Its way too stiff: because of the (3) #6AWG (hot/hot/neutral) and the (1) #8AWG conductors, the cable is almost an inch in diameter, and is very stiff. It does not uncoil well, and it's difficult to get the full length out of it, without a helper to assist stretching it out.

I was unhappy with it, so I returned it to Amazon. Thinking about a solution, I started browsing eBay one day, and came up with a 25' length of 6/3 SOOW cord for $68, best of all this cord weighs 568 pound per thousand feet (or .568 pound per foot), this 25' cordage only weighed a little over 14 pounds. This is about 1/2 the weight of the above 30' 6/3+1/8 14-50 cord from Amazon. Good so far.

I decided to use 6-50 plugs and receptacles, because the UMC has that as an option, also when installing 6-50's in "the wild" at distant friends houses, 6/2 is much more economical to buy than 6/3, we only need 240V + ground, no need for a neutral. So I then bought 6-50 receptacles and plugs from Amazon, along with a 14-50 Male, this so I could make a short "adapter" cable from a 14-50 to a 6-50 receptacle, to use the extension cord, or plug the UMC directly in with the 6-50 adapter attached to it, this is so the cord can be used at RV Parks, if needed.

Here are the links:

25' 6/2 SOOW cordage from eBay, $68 with shipping included:
6 3 SOOW So Cord 25 ft HD USA Portable Outdoor Indoor 600 V Flexible Wire Cable | eBay

I also bought 2' from the same seller for the 14-50 Male to 6-50 Female adapter cord
6 3 SOOW So Cord per Foot USA Portable Outdoor Indoor 600 V Flexible Wire Cable | eBay

Then buy (2) 6-50 Receptaces: $11.65 each
Leviton 5378 50 Amp, 250 Volt, Surface Mounting Receptacle, Straight Blade, Industrial Grade, Grounding, Black - Amazon.com

One 6-50 Plug: $16.55 (Home Depot has it for about $15 as well)
Leviton 931 30/50 Amp, 250 Volt, Plug, Angle, Grounding, Black - Amazon.com

The add in one 14-50 for the adapter (think RV parks, you might need an extension), $14.42
Camco 55255 RV 50 Amp PowerGrip Replacement Plug : Amazon.com : Automotive

Here is the cord, l put together, 14-50 to 6-50 adapter in the 2nd picture. You might not save much money on building this extension cord, but it's lighter and far easier to unroll and use than the CAMCO 14-50 extension cord...

View attachment 20298

View attachment 20299

Thank you for putting this up. I followed exactly these steps and made a 6/3 extension with 10-30 Dryer outlet.

 

[ManjuM]

On the CAMCO 14-50, I just did not insert the neutral blade, is has no purpose when used as an adapter on a 6-50, since a 6-50 is just 2 hots and the ground. I think you can get away with 50 feet or so, but you should probably build it as 2 cords, maybe (2) 25 footers, or a 15' and a 35', as it's always better to use the shortest possible extension that works.

To build a 6-50 extension cord, you only need 6/3 SOOW, if you are building a 14-50 extension cord, you would need 6/4. (For the neutral conductor).

Personally I wanted the lowest weight cord, that's why I went with the 6-50 and built the RV Park 14-50 to 6-50 short adapter cable; I can still use a 14-50 in the wild, I just don't need to carry %25 more weight on the extension cord cabling... I also bought the matching 6-50 UMC adapter as well, of course.

For the short adapters you make up, stay with 6/3 as well, you want to keep the same size wire, 6/3 SOOW I believe is rated for either 45 or 55 amps with 2 current carrying conductors, I seem to remember.

Edit: the eBay auction states 45A per conductor on the 6/3 SOOW link:
6 3 SOOW So Cord 25 ft HD USA Portable Outdoor Indoor 600 V Flexible Wire Cable | eBay

Totally agree with you. I had the same though. For 1 6/4 is too bulky and you dont need the 4th wire anyway. I made a 100"ft 6/3 6-50 extension and a short 2"ft 10-30 dryer to 6-50R to connect to dryer outlet.