Hello Michael, Just leave the 4 gage in place and install another 4 gage in parallel with the existing wire. We do this all the time in old and new installations. Two 4 gages in parallel will run cooler than one 1 awg cable because of more surface area.
If you installing these wires in a conduit such as a water proof flexible plastic or metal conduit, it is best that the conduit is 60 percent larger than the wire to allow air flow. Do not seal the end of the conduits, unless it is connected with seal tight box connectors to a ventilated enclosure. The current capacity of the wire should at least 1.25% larger than the continuous current in the wire. Example: If you draw 100 amps continuous, then a wire good for 125 amps should be used. Two 4 gage wires with a temperature rating of 75C is about 80 amps per wire or 160 amps total for a standard insulation wire in conduit. Not in conduit suspended in air for 100 feet is closer to 200 amps allowing for a 3 percent voltage drop. Roland ----- Original Message ----- From: Michael K Johnson<mailto:[email protected]> To: [email protected]<mailto:[email protected]> Sent: Saturday, November 02, 2013 9:06 AM Subject: [EVDL] Wire gauge for 48V lawn tractor/mower conversion: sanitycheck I'm new to the list. I expect this has been discussed before but I haven't found the right search terms. I apologize if so and would appreciate a pointer. I'm also entirely new to EV conversions. I've been scouring the net, and have purchased most of the items to convert my lawn tractor to an EV this winter. I'm using the ME1004 as a "drop-in" replacement for the ICE in my existing hydrostatic-drive donor tractor, not doing separate deck motors. I know several of the things I'll want to do to reduce loss (belts, bearings, lube, new sharp blades, etc.) and I understand the single motor/multi motor tradeoffs and already have the ME1004... ☺ From what I've seen so far, ME1004 conversions on lawn tractors at 48V consume 70-100 amps while actually mowing, and may momentarily consume up to 200 amps while spinning up the mower deck. I misread something somewhere (I don't even remember where anymore) as indicating that since I expect to have 20 feet or shorter, I could use 4AWG fine-strand welding cable, so I bought some. Then I learned that this might have been somewhat optimistic, so I'm expecting to chalk that up in the "mistakes" column. I found the helpful articles at engineeringtoolbox.com and a few references on resistance of copper wire and am trying to calculate real voltage drop instead of following rules of thumb. In particular, I'm looking at http://en.wikipedia.org/wiki/American_wire_gauge<http://en.wikipedia.org/wiki/American_wire_gauge> and http://www.engineeringtoolbox.com/copper-wire-d_1429.html<http://www.engineeringtoolbox.com/copper-wire-d_1429.html> which both have similar numbers for ohms per thousand feet for copper wire. I intend to fuse the conversion no larger than 400 amps and possibly at 200 amps using an ANL fuse. So I've been calculating voltage drop over 20 feet (and also, pessimistically, at 30 feet in case it takes more cable than I expect) at 100 amps, 200 amps, and 400 amps. My understanding is that I want to keep the voltage drop below 2%, so with some room for error it seems like I want less than 0.9V drop in normal operation, and not go much above it momentarily. I'd like a sanity-check on my math, as well as the wisdom of the list on recommended wire gauge... For resistance per Kft, I am using: 1/0 awg: 0.09827 1 awg: 0.1239 4 awg: 0.2485 (I see no point in buying smaller than 1 awg if I replace the 4awg I bought, so I'm ignoring 2 awg in my calculations.) It looks to me like the voltage drop per 10 feet at 100 amps is the same as ohms per Kft, since I divide by 100 to get the resistance of 10 feet, then multiply by 100 amps to get the voltage drop, so it cancels out. Multiplying by 2 should give me voltage drop at 20 feet at 100 A (my expected normal operation) and by 8 should give my voltage drop at 20 feet at 400A. Since 1/0 awg is more than twice as expensive as 1 awg as well as harder to work, I'd prefer to use 1 awg. I calculate that 1 awg (0.1239 Ohms/Kft) should drop about 1/4V at 100A at 20 feet (max normal load) and about 1V at 400A at 20 feet (max momentary load). Am I missing anything? Thanks much! _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub<http://www.evdl.org/help/index.html#usub> http://lists.evdl.org/listinfo.cgi/ev-evdl.org<http://lists.evdl.org/listinfo.cgi/ev-evdl.org> For EV drag racing discussion, please use NEDRA (http://groups.yahoo.com/group/NEDRA<http://groups.yahoo.com/group/NEDRA>) -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.evdl.org/private.cgi/ev-evdl.org/attachments/20131102/9df4410d/attachment.htm> _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org For EV drag racing discussion, please use NEDRA (http://groups.yahoo.com/group/NEDRA)
