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WIRING YOUR BOAT ----> READ THIS FIRST

 
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merwin10
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Joined: 11 Oct 2006
Posts: 440
Location: Buzzards Bay, Massachusetts

PostPosted: Wed Mar 19, 2008 3:44 pm    Post subject: WIRING YOUR BOAT ----> READ THIS FIRST Reply with quote

Hi Guys and Gals -

I have been asked a number of times why do I have such a low voltage on the bridge when the instruments are all on - The reason is that your wiring is ampage starved. Below is the simple explaination why and what to do about it -

=============================================

High Current DC Boat Wiring

________________________________________
The issue here is high current wiring needs at the helm some distance from the where the batteries are. Often boat manufacturers run a couple of extra circuits to the helm for DC powered devices. The issue with us with old boats is that there was not that much in the accessory area available when the boat was built. The circuits if any are inadequate for many of today’s devices and will exceed the capacity of these the wire capacity in the circuits that are there.

Two of the high current devices we install today on our boats are stereo sound systems and radar. My boat has a nice 10AWG auxiliary circuit to the helm. One might at first believe that 10 AWG will handle 50 amps this should be sufficient to run any high current system, however it is marginal at best due to the distance from the battery and the subsequent voltage drop.

Here is the Issue

When powering high current devices via DC, there is a significant problem concerning voltage drop along the wire based on the diameter of the wire and the distance from the battery and the amount of ampage needed. The DC voltage drop can be so severe that it must be accounted for anytime you power high-current devices.
Most of you may have experienced voltage drop related issues with accessories after you wired a new circuit. Even today with all the standards, boat manufactures tend to not specify the voltage drop! Most boats just are not wired to handle high current DC loads.

The ABYC breaks accessories down into two groups 3% and 10% devices. The 3% devices are those that are critical to safe navigation. The rest fall into the 10% group! The issue is many devices are not specified. One might say well lights must be in the 10% group. Not really, it depends on the use of the light! Yes the salon lights might be in that 10% group but your navigation and instrument lights are not!

Other considerations are sensitive devices like a Stereo sound system fall in the 3% drop group. No it is not a critical device but the power needs to be stable and plentiful for proper operation.

Our scenario say we have 12VDC at the battery - then a 3% voltage drop means that the voltage found at the electronic device should never be less than 11.64V. So where did the 3% loss in voltage happen? Well as you know most things have a resistance to flow it is this internal resistance in the wire itself where the loss occurs. The amount of resistance depends on the wire size, current flow, and distance of the run. While we are talking about the distance in electrical terms it is the physical distance times two, one for the positive side and one for the return or negative side. More on this later!



So looking at the drawing above most see a battery wire and radar unit. In electrical terms it is a simple series resistive circuit with three resistors and a battery. The electrical schematic looks like what you see below. Normally, the resistor values of R1 and R3 are insignificant, until you deal with high current devices. Only then do these resistances become significant enough to lower the voltage at the load to an unacceptable level.



So what can we do to insure that the current across R3 (the radar unit) is sufficient and within the 3% voltage drop- We could -

• Increase the source voltage at the battery.
• Decrease the wiring length.
• Decrease the load requirements.
• Increase the wire size.

Looking at the above solutions only one is possible, increase the wire size.

So what is the increase wire size??

Electrical engineers use a formula to determine the correct size wire. Before we go much further I will tell you that there are two standards of wire SAE and AWG. In the marine environment we only use Tinned AWG strained wire. If you use SAE wire you will need to increase the diameter even further as it has a different diameter specification. Un fortunately in our old boats most of the wiring is SAE copper and is not well suited for the marine environment. As copper corrodes it becomes more resistive to current flow, until it breaks down completely.

The formula we use is below -

CM = (K x I x L) / E

Where:

CM = circular mils of the wire.
K = 10.75 (Constant representing the mil-foot resistance of copper)
I = Current (amps)
L = Length (feet)
E = Voltage drop (in volts)

Remember when figuring Length it is the physical length times two. Things like cars and truck are made of metal hence the whole vehicle chassis is the return path, negative side of the battery. Where our old boats are made of wood or fiberglass there is no such chassis so you need to account for the return path, negative side of the battery.

I want to power my radar unit, which includes a chartplotter, radome, and GPS unit by a single power cable. It requires up to 10Amps @ 12VDC.

I want to see if the existing factory-provided 10AWG circuit is sufficient for this purpose. I have determined that there are 18 cable feet from the main distribution panel to the helm area where the circuit is terminated. To get the round trip distance, I must multiply this by 2, so I need to use 36ft.

I next need to find out how much current that 10AWG can carry but still provide less than a 3% voltage drop, so I now start plugging values into the formula.

There are 9,343 circular mils in a 10AWG wire. More than likely your boat has 10 SAE so the voltage drop will be greater.

So, to solve for voltage, the formula is re-arranged as follows:

E = (K x I x L) / CM

10.75 x 10 x 36 = 3,870

3,870 / 9343 = 0.41V

Unfortunately, 0.41V is more than a 3% voltage drop (3% = 0.36V). We will have to use a larger gauge wire. Of course now would be a good time to consider any other devices that you may want on the bridge such as a vhf radio, fish finder and etc. A rule of thumb is always throw in an added 20% amps for future needs. Or if you find your calculation is close to the next wire size go up one size.

Now that you are totally confused as 10 gauge wire will handle 50 amps of current, this may indeed be true, but not with the limit of a 3% voltage drop. Lets use the formula again to see just how much of a voltage drop 50 Amps will produce on that 36 feet 10AWG wire:

10.75 x 50 x 36 = 19,350

19,350 / 9,343 = 2.07V

Therefore, a 12V powered circuit will have 9.93V at the instrument, which likely will prevent proper operation or operating at all since most units today have voltage shut downs that will force the unit off if the voltage is to low or high.

What you are looking at is the 50Amp rating is a safety rating, and specifies the maximum amount of current that can safely flow within the wiring under ideal conditions. The 50 amp safety rating has nothing to do with the amount of voltage drop across the wire at high current flow rate. While we are talking about safety rating s you should know that these ratings are at 70 degrees F. Raise the temperature and the ampage safety rating goes down. So in an engine room where the temperature may be 180 degrees the 10 SAE wire needs to be derated to 43 amps.

It is interesting to note however, that same 10AWG wire be used with a 120VAC circuit, could safely carry 30Amps without any significant voltage loss. Voltage drop problems occur with high-current DC powered devices only.

The Solution!!!!

To meet my needs, I will use the next larger wire for the helm wiring; 8AWG. I have determined that 8AWG has 14,810 Circular Mils. If you recall, 10AWG wire had 9,343 Circular Mils. Therefore, 8AWG is 60% larger than 10AWG. So since I am using 8 AWG wire what is the current capacity for 36 feet of length with a 3% voltage drop.

I = (E x CM) / (K x L)

(0.36 x 14,810) / (10.75 x 36)

The answer: 13.8Amps

This will be enough to allow me to power the GPS/Radar/Chart plotter and other miscellaneous items. I can also power the VHF radio either from this circuit, or the 10AWG circuit that is already installed, which will help "split the load".

So why go thru all these calculations well wire is expensive at today’s cost of $2.60 per foot 18 feet of red and 18 feet of black is about $100, so this is not an inexpensive venture.

What you should be getting out of this is that distance is a current killer. Since the boat has a flybridge, a much longer run is required than if the boat were simply an express cruiser where the helm was only a few feet away.

Chart of Current in Amps versus distance in feet



Here are the formulas for figuring Circular Mils, Current, Length and voltage drop

CM = (K x I x L) / E

I = (E x CM) / (K x L)

L = (E x CM) / (K x I)

E = (K x I x L) / CM


where

CM = Circular Area of Conductors
K = 10.75 (Constant representing the mil-foot resistance of copper)
I = Current (amps)
L = Length (feet)
E = Voltage drop (in volts)

Circular Mils (1 mil = 0.001" diameter):
Gauge Circular Mils
18 AWG = 1,537
16 AWG = 2,336
14 AWG = 3,702
12 AWG = 5,833
10 AWG = 9,343
8 AWG = 14,810
6 AWG = 24,538
4 AWG = 37,360
2 AWG = 62,450
1 AWG = 77,790
0 AWG = 98,980
00 AWG = 125,100
000 AWG = 158,600
0000 AWG = 205,500

Note: Hope this was of some help to some of the wiring issues you are facing!

Mike - Wink
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