Marine grade wiring

[Bubbasboat]

I am unfamiliar with marine grade wire. I know it is multi strand tinned for corrosion resistance. What confuses me is the amperage rating. I see for instance charts (see Ref. 1 below) which say 14 AWG is rated for 20 amps DC. I would never consider using a 14 AWG wire for anything greater than 15 amps. When I look up 14 AWG marine grade to buy, I see it is rated to 35 amps outside engine compartment (see ref. #2 below). I'm confused, can anyone help?

I am running a 20 amp 12v dc power jack to my dash. Length of circuit one way is approximately 12 ft. Should I use a 14 AWG or a 12 AWG for this given non critical circuit?

I am also running a seperate 14 AWG for a dual 5v, 1 and 2.1 amp usb power charge ports. Could I supply these of the same wire as the 12 vdc 20amp power plug supply?

Are there any other differences of interest?

Ref. 1


Ref. http://www.bluesea.com/articles/1437

Ref. 2
"14 AWG marine grade primary wire features flexible, tinned copper stranding which is ideal for your boat wiring needs. Our marine wire is made in the USA and meets ABYC and US Coast Guard specifications while UL listed as UL1426 Boat Cable. The maximum allowable amperage of single conductor 14 AWG marine grade primary wire is 35 amps outside of engine spaces and 30 amps inside the engine space"

Ref. http://www.bestboatwire.com/14-awg-p...pper-red-25-ft

 

[kjsAZ]

The specified current is the rating the wire can handle without exceeding the max temperature which would melt the insulation. However, that's at room temperature which is why the engine compartment is excluded. What isn't included is the voltage drop you will get. The wire has resistance and depending on the current you will get a lower voltage to your load. The tables are usually based on some percentage of voltage drop (common are 2.5%, 3% and 6%).
There are "consumers" which run a lot better with higher voltage (i.e. starter or your VHF = more power out) and you want to keep the voltage drop low. Instead of the tables I would use one of these calculators where you can enter the voltage drop you want to allow plus the length of positive and negative wire combined and they will give you the required AWG.
Thicker wire is always better......

 

[jhebert]

In any low-voltage power distribution system, there is more concern for the influence of voltage drop in a conductor distributing electrical power than for the current handling rating. Any rating of wire for current alone is figured based on the allowed heat rise. The allowed heat rise is a function of the enclosure surrounding the wire. Wire run in bundles will be rated lower than a single wire in free air. The insulation on the wire also affects heat rise and current rating. You should completely ignore all of that. Heat rise is of no value in figuring conductor size in 12-Volt systems for any length of wire beyond a few feet.

The simplest way to approach the current capacity of a conductor for 12-Volt power distribution is to figure its current rating is based on the more or less standard computation for 1-Ampere per 700-mills circular cross section. This means the current rating for wire gauge looks like this:

00-AWG = 190-Amperes
0-AWG = 150-Amperes
1-AWG = 119-Amperes
2-AWG = 94-Amperes
3-AWG = 75-Amperes
4-AWG = 60-Amperes
5-AWG = 47-Amperes
6-AWG = 37-Amperes
7-AWG = 30-Amperes
8-AWG = 24-Amperes
9-AWG = 19-Amperes
10-AWG =15-Amperes
12-AWG = 9.3-Amperes
14-AWG = 5.9-Amperes
16-AWG =3.7-Amperes

Now we consider voltage drop. There is a very simple rule of thumb--I call it Hebert's Rule because I believe I am the first person to propose this rule.

The maximum length of a conductor that can be used for two-way power distribution is equal to the system voltage if the current is to be at the 1-Ampere per 700-circular-mills rated current of that conductor and voltage drop is to be three-percent or less.

In other words, in a 12-Volt system of power distribution, the maximum distance for power distribution at the rated current is 12-feet.
For example, if you want to distribute 15-Amperes over a distance of 12-feet and not have more than three-percent voltage drop, the conductor must be 10-AWG.

I don't know on what authority someone told you that you could use a conductor of 14-AWG to distribute 35-Amperes of 12-Volt power. That is nonsense. Just for 12-feet of length, you need 6-AWG conductors to deliver 35-Amperes of 12-Volt power with less than three percent voltage drop.

You can scale the current to longer lengths in a linear manner. For example, if you want to deliver power over 24-feet instead of 12-feet, you have to double the current rating of the conductor. Using the 35-Ampere example, to deliver at 24-feet with three-percent drop means using 3-AWG conductors.

In your particular case, running 20-Amperes to 12-feet, the power distribution conductor must be 8-AWG if you want to maintain a three-percent or less voltage drop. And, believe me, you certainly do want to have only a three-percent or less voltage drop.

 

[Bubbasboat]

The 35 amp rating question was answered "The specified current is the rating the wire can handle without exceeding the max temperature which would melt the insulation" by kjsAZ.

My interest is in a 10 and 20 amp circuits, about 12 ft at most.

I used one of those calculators. One for my 10 amp and one for my 20 amp. Seems that a 12awg would work fine for both. Your list above seems to indicate I would need an 8awg to handle the 20 amp. I'm still confused.

These are non critical items, so the power loss should be acceptable.

Screen shots of calculator results below.

Followed by a photo of the devices I'm installing and their requirements per packaging.









Is this right or not?

 

[bruceb58]

Very few circuits on a boat would require a 3% loss limit. Most things could probably go as low as 10%. Depends on what you are powering.

If you can live with just less than 10% voltage drop and you are running 12' of positive and 12' of ground, you can use 14 AWG. 12 AWG would give you 6% which is likely just fine for 95% of things you will be plugging into that outlet.

 

[Bubbasboat]

Very few circuits on a boat would require a 3% loss limit. Most things could probably go as low as 10%. Depends on what you are powering.

If you can live with just less than 10% voltage drop and you are running 12' of positive and 12' of ground, you can use 14 AWG. 12 AWG would give you 6% which is likely just fine for 95% of things you will be plugging into that outlet.

Thank you. Getting less confused all the time!

The only things I will be using is a GPS in the power socket and a cell phone in the 2.1amp usb socket.

If I use 12 AWG on the on the power socket, can I supply the USB off same wire or should I run a second supply for the USB?

 

[kjsAZ]

It depends (as usual). If you plan to connect equipment to the power outlet which can generate spikes (like a motor or other high current equipment) it would be better to have independent wires for plus and minus. For what you plan to use it for a single set should be sufficient but make sure it's sized and fused for the combined current.
Remember, the fuse should be as close as possible to the battery as it then also protects the wire.

 

[Bubbasboat]

It depends (as usual). If you plan to connect equipment to the power outlet which can generate spikes (like a motor or other high current equipment) it would be better to have independent wires for plus and minus. For what you plan to use it for a single set should be sufficient but make sure it's sized and fused for the combined current.
Remember, the fuse should be as close as possible to the battery as it then also protects the wire.

Well, given that thought about spikes and usage, one never really knows for sure what will happen later, so I will install a separate +/- for each item.. A 12awg for the power socket and a 14awg for the USB ports.

Thanks for everyone's help.

 

[kjsAZ]

Good choice. Not much difference between getting two or four wires around the boat.

 

[jhebert]

...I will install a separate +/- for each item.

You are welcome to take that approach, that is, to distribute power on the boat by wiring every circuit directly to the battery terminals, but it is a bad idea. It is only endorsed by people who lack understanding of electrical circuitry. But, again, feel free to follow bad advice, as you got it for nothing.

 

[bruceb58]

You are welcome to take that approach, that is, to distribute power on the boat by wiring every circuit directly to the battery terminals, but it is a bad idea. It is only endorsed by people who lack understanding of electrical circuitry. But, again, feel free to follow bad advice, as you got it for nothing.

Not sure where he said he was going to wire directly to the battery.