Community Forums

GyroGypsy · 08-26-2019, 12:32 PM

How difficult is it to pull larger wires through the coach looms on a WBGO N24?

Can I just attach new wires and pull? The solar charge controller location above refrigerator makes a long path to the coach batteries, and I'd like to upgrade from existing #8 wires.

John Hilley · 08-26-2019, 11:13 PM

I doubt you can pull it through the bundle. It is probably secured in several places.

GyroGypsy · 08-27-2019, 05:32 PM

I was hoping to find somebody with experience. It outwardly looks like flexible conduits are used, so pulling through should not be "too difficult" (yeah). This impacts my decision to use parallel rather than series solar panel connections.

ALLOY · 08-31-2019, 10:13 PM

Filling loom/conduit too much reduces the capacity of the wire.

GyroGypsy · 09-01-2019, 06:41 AM

Quote:

Originally Posted by ALLOY

Filling loom/conduit too much reduces the capacity of the wire.

What do you mean?

**Spdracr39** · 09-01-2019, 06:55 AM

Theoretically it increases the amount of heat and thus increases the resistance of the wires. In reality I don't think it would make a lot of difference or they wouldn't bundle them together in the first place.

ALLOY · 09-01-2019, 08:43 AM

Quote:

Originally Posted by spdracr39

Theoretically it increases the amount of heat and thus increases the resistance of the wires. In reality I don't think it would make a lot of difference or they wouldn't bundle them together in the first place.

Don't expect anything but the absolute miminum (or less) when it comes to RVs.

Here's a fill chart

http://conduitfillcalculator.com/conduit-fill-chart/

GyroGypsy · 09-02-2019, 04:44 PM

Let's start with 20 ft of #8 AWG wire, which has a total resistance of .6282 * (20/1000) = 0.0125 ohms. The Temperature Coefficient of Copper is .39 % per degree C. Let's heat it up from 20 to 30C (68F to 86F), and we will find the resistance will increase by (0.0039 * 10) ohms and become 0.0125 + 0.039 = 0.0515 ohms. If we were running 15A through that #8 wire, then the difference would be [E=I*R] (15 * .0515) - (15 * 0.0125) = .59 volt additional loss due to higher temp. If on a 12 volt system, this is .59/12 = 4.9% change due to temperature. Hot stuff!

I found the physical # wires limit also interesting.