Wiring, wire size, and current capacity

[jlinder]

I would like to get some opinions on this. I am doing maintenance/upgrades on my boat, some of which is electrical. I have been trying to study up on all the marine standards in order to avoid mistakes. I have found tons of rules and standards for safety, and most of the new ones for me concern the need to prevent any kind of spark that would ignite gas fumes.But with all the strict safety precautions I was completly suprised when I found out the specs for wire gauge and current capacity.I found a site listing ABYC standards ( http://www.pkys.com/Reference.htm ) which tells me that I can run 20 amps through an 18ga. wire. This was very suprising in that electrical codes (for dry land) limit you to 7.5 amps to prevent overheating! Even worse is the voltage drop you get if you actually try to run 20 amps through 18ga.He comes the math. I looked up the standard resistance for 18ga copper wire and got 6.6 ohms / 1000 feet ( http://www.bnoack.com/index.html?htt...esistance.html ), or .0066 ohms/ foot.Suppose I have something that draws 20 amps that has 10 feet of wire from the fuse block, and it is 15 feet of wire from the fuse block to the battery. (Not hard on a 20 foot boat).Total of 25 feet of wire, but you have to consider the path from the battery to the device and back, for a total of 50 feet of wire. Total resistance is .33 ohms (doesn't sound like much yet, does it?)Voltage drop = resistance x current.Voltage drop = .3 x 20 ampsVoltage drop = 6.6 voltsVoltage to device = 12.6 volts from battery - 6.6 volt dropVoltage to device = 6 volts Wattage dissapated in wire = 6.6 volts x 20 ampsWattage dissapated in wire = 132 watts.Don't know about you, but trying to run a 12 volt device with only 6 volts and dissapating 132 watts in a wire is a BIG NO-NO in my book.And remember, this assumes all you connections and fuses are perfect!!!Please, I would like some comments. (But I will be running heavier cable)

[sangerwaker]

Jack,I really had a hard time believing the chart on the link you gave, so I went in search of another chart. That appears to be an accurate chart according to several websites I looked at. Personally, I would never try to run 20 amps through an 18 AWG wire either.I sell electrical goodies (AC type stuff) to contractors, and 12 AWG is the norm for 20 amps at 120V.I rewired my old boat also, and as I have just learned, I WAY over-engineered the job. Looks I went way overkill on the wire guage. Oh well, it's not like the wire is really expensive, and I would certainly rather be safe than sorry!Here is one of the most complete pages I found. web page

[18rabbit]

Originally posted by Jack L:Voltage drop = .3 x 20 amps

Voltage drop of 3%??? Your math should be:Voltage drop = .03 x 20 ampsFeel better about it now? Btw, to get 20-amps 25-ft out there and 25-ft back (50-ft total) you want to use AWG 4 conductor for a 3% drop in voltage, AWG 10 conductor for a 10% drop in voltage. Note: you would use .1 in your formula when calculating for 10%. Also, in all those specs and codes you should have read where AWG 16 is the smallest wire you will be using on your boat. You should not have any AWG 18*, with or without 20-amps. Fwiw, AWG 18 will move 5-amps, 12vdc, 10-ft total. Note: a 24v system needs AWG 14 to move 20-amps a total of 10-ft; a 12vdc system needs AWG 10 to move 20-amps 10-ft. As you are aware, you will need bigger conductors for less voltage.If you are comparing amps-AC on terra firma with amps-DC on your boat you will likely get frustrated. AC does not have as sever a voltage drop issue as DC current does. Also, you can cram more amps-AC into a conductor than you can amps-DC.*EDIT: technically, AWG 18 is ABYC approved for use but only if it is a small jumper that is entirely enclosed within a panel.

[Paul Moir]

Jack - your reading the standard in a vaccum. Consider this:ABYC E-9.15.8 Conductors used for panelboard or switchboard main feeders, bilge blowers, electronic equipment, navigation lights, and other circuits where voltage drop must be kept to a minimum, shall be sized for a voltage drop not to exceed 3 percent. Conductors used for lighting, other than navigation lights, and other circuits where voltage drop is not critical, shall be sized for a voltage drop not to exceed 10 percent.So you could never be allowed to install 18 guage wire in a circuit where you had a voltage drop of 50%.On the same page as you posted, click on 'Excerpts from Standards'.

[crab bait]

not doubtin' that you read what you read.. but it's not right an is wrong..big diff from ac to dc an what is lackin' from your 'scientific' calculations is voltage.. the more volts ,,the smaller the wire.. the less volts ,, the bigger the wire.. 12 volts is about as 'less' as it gets..ever think about becomin' an electrical engineer..

[Ralph 123]

Just to reiterate Rabbit's point:

Technical Data - American Boat & Yacht Council Standards for Boats Conductors shall be at least 16 AWG (except 18 AWG may be used as internal wiring in panelboards). Conductors shall have a minimum rating of 600 volts (E8.14.1.1). All conductors and flexible cords shall meet the requirements of the applicable standards of Underwriters Laboratories, Inc. (E8.14.1.6). Flexible cords shall have a minimum of 300 volts (E8.14.1.2).

Here is a link to a good wires size calculator: http://www.ancorproducts.com/Technic...calculator.htm Here is a link to a good summary of the standards: http://www.ancorproducts.com/Technic...Standards.html

[Ralph 123]

Here is a good general discussion of boat wiring by Don Casey: http://www.boatus.com/boattech/casey/05.htm

[jlinder]

Got a good thread going here - thanks for the replies.sangerwaker -One thing to note is that with 120v stuff there is not only a maximum current for a wire size but they want a maximum voltage drop of 2% (if I remember correctly). If the wire is long enough you have to up the gauge. Trying to run gear at an undervoltage is one of the fastest ways to need replacements. It is really painful to see a large electric motor try to start when someone failed to take motor starting currents into account 18 rabbit - You read it wrong. I never used percentage. the .3 (typo - should have been .33) is the resistance in ohms of the wire. If you put 20 amps across .33 ohms you get 6.6 volts of voltage drop across the wire.Also, the voltage drop problem is much bigger with 12 volt systems than with 120 volt since you are concened with the percentage drop, and 3% of 12 volt is much smaller than 3% of 120 volt.I do not believe that AC vs. DC is a difference here. I think what you mean is 120v systems vs 12v systems. (For those of you who are sticklers for accuracy there is a difference due to what is called the skin effect but it does not become significant until the frequency becomes 1MHz or more).Paul Moir - Great point. My concen is that while you and I know to consider voltage drop, how many other people do. I doubt very many have even heard of it.It may have just been a stupid oversight on my part, but voltage drop limitations do not seem to be very evident when I look at the web sites. You had to click on "Excerpts from Standards". From what I can see this "excerpt" is a limiting factor more often in wire gauge than total amperage.Would love to see it more out front.crab bait - We think of the more volts the smaller the wire because to deliver the same power if you double the voltage you half the current. At home you use 240volt for air conditioners, ovens, etc. to keep the wire gauge reasonable.But that is not what is happening here. We started with a fixed current (20 amps in our example) and calculate the wire needed due to that (and percentage of voltage drop).BTW - if you think 12 is as less as you get you should see a 300 amp 5 volt system. And I have worked on designs of systems from microvolts (receivers) to 20 kilovolt, from frequencies from DC to 40GHz.Thanks everyone so far, hope we get more interest - I think this is a topic that may boat owners don't fully understand.

[Ralph 123]

It may have just been a stupid oversight on my part, but voltage drop limitations do not seem to be very evident when I look at the web sites.

It is usually pretty prominent. You'll often find a wire calculator or size chart whenever the subject is covered. One way to think of the Allowable Amperage ratings is a max capacity rating below which the wire will not begin to heat to the point of setting the insulation on fire.Consider this caveat for the very same table from Ancor Marine:

The values shown on the ampacity table are the maximum safe amperages which the conductor can carry on a continuous basis. They do not apply to intermittent starting loads such as motor start currents.

http://www.ancorproducts.com/Technic..._Amperage.html

[ron7000]

Originally posted by Jack L: I would like to get some opinions on this. I am doing maintenance/upgrades on my boat, some of which is electrical. I have been trying to study up on all the marine standards in order to avoid mistakes. I have found tons of rules and standards for safety, and most of the new ones for me concern the need to prevent any kind of spark that would ignite gas fumes.But with all the strict safety precautions I was completly suprised when I found out the specs for wire gauge and current capacity.I found a site listing ABYC standards ( http://www.pkys.com/Reference.htm ) which tells me that I can run 20 amps through an 18ga. wire. This was very suprising in that electrical codes (for dry land) limit you to 7.5 amps to prevent overheating! Even worse is the voltage drop you get if you actually try to run 20 amps through 18ga.He comes the math. I looked up the standard resistance for 18ga copper wire and got 6.6 ohms / 1000 feet...Suppose I have something that draws 20 amps that has 10 feet of wire from the fuse block, and it is 15 feet of wire from the fuse block to the battery. (Not hard on a 20 foot boat).Total of 25 feet of wire, but you have to consider the path from the battery to the device and back, for a total of 50 feet of wire. Total resistance is .33 ohms (doesn't sound like much yet, does it?)Voltage drop = resistance x current.Voltage drop = .3 x 20 ampsVoltage drop = 6.6 voltsVoltage to device = 12.6 volts from battery - 6.6 volt dropVoltage to device = 6 volts Wattage dissapated in wire = 6.6 volts x 20 ampsWattage dissapated in wire = 132 watts.

Hi, was reading your inital post and I got a little confused by your math and how you came to your answers. I think you are incorrect, unless your accouting for something else or I misunderstood you.But, if you have a device "that will pull 20 amps", then that is because it has a certain resistance at a certain voltage. For the argument here, I'll use 13.0 volts, about halfway between a battery at rest at 12.0-12.5V and when the alternator is putting out anywhere between 13-14.6V. Everything is based of V = I * R. So if a device pulls 20 amps at 13.0 volts, then it's R (resistance) must be 0.65 ohms.Now go back into the wiring diagram of your circuit, with wire resistance at 6.6ohms per 1000 ft.{13.0V} + ---- [Rw1 (15ft)] -- [Rbus] --- [Rw2 (10ft)] --- [Rdevice = 0.65ohms] ---- [Rw3 (25ft)]-- Ground.Rw1 = resistance of 0.099 ohms of 15 ft wireRbus = guesstimated resistance of bus at 0.05 ohms?Rw2 = 0.066 ohms for 10ft wireRdevice = must be 0.65 ohms to pull 20 amps at 13 volts.Rw3 = 0.165 ohms for 25ft wire from device back to ground.Resistance Total in circuit is all R's added together = 0.964 ohms Correction => 1.03 ohmsIn this series circuit, current is constant from + post of battery to - post of battery, using V = IR, I = 13.0 / 1.03 = 12.621365 Ampsyou will have (rounded off) 12.6 amps flowing through the 18 guage wire to your device and back to ground (neg post of battery). The device, that is supposed to pull 20 amps at 13V, can only pull 12.6 amps because of the extra resistance in the circuit. So now go back, using V=IR at each R value to determine the voltage drop and find out the voltage being supplied to the device:V Rw1(15ft) = 12.621 amps * .099 ohms = 1.2495 VV Rbus = 12.621 amps * .05 ohms= 0.63107 VV Rw2(10ft) = 12.621 amps * .066 ohms = 0.8330 VVoltage at device = 13.0 - 1.2495{Rw1} - 0.63107{Rbus} - 0.8330{Rw2} = 10.2864 V voltage drop across device is V = I*R = 12.621 amps * 0.65ohms = 8.20388 voltsvoltage drop then, across Rw3 (wire to neg post of battery) must be the remaining voltage left to drop to zero. Typically, in circuit analysis the resistance in wire is negligible and assumed to be zero. So the voltage on the wire going from the negative side of the load to the ground is always zero! However in this scenario we are accounting for the wire resistance!V drop of Rw3 = 12.621 amps * 0.165 ohms = 2.0825 Voltssubtraction of all volt drops in circuit (Rw1, Rbus, Rw2, Rdevice, Rw3) between voltage source and ground must equal zero. That's one of the golden rules of electricity. Source voltage here was 13.0V so:13.0 - 1.2495 - 0.63107 - 0.8330 - 8.20388 - 2.0825 = 0.00005 volts close enough Not as bad as 6V like you initially said. Basic idea behind wire sizing is the more current going through the wire, the greater the voltage drop over the length of it, so you either want to reduce the amount of current going through it or reduce the length. If you ran 20 amps through 2ft of wire at 6.6ohms/1000ft R, then voltage drop is only 0.264V, and the power dissipated there is I^2*R = 20A * 20A * 0.0132 = 5.28W. Then if the Volt drop is acceptable, you get into how much heat a type of wire can handle, which is based on conductor and insulation material.Wattage Rw1 = 15.77, over 15ft of wire remember so heat isn't localized;Wattage Rw2 = 10.51Wattage device = 103.5 watts at 12.6 amps; should be 260 watts at your spec of 20amps at 13volts. Calculate this by either 12.621 amps * 12.621 amps* 0.65 ohms, or by 8.20388 voltdrop * 12.621 amps.Formuals for power P (watts) is P = V * I P = I*I*R P = (V * V) / Rfor something like a length of wire, you need to use the I^2/R because voltage varies (drops) over the length. It's the current that remains constant.18awg stranded is great on small runs, like instrument to instrument behind a console because it's thin and the runs rarely exceed a foot and you're never pulling over 10amps. For the runs from the motor area to the console to a switch then off to an instrument, 14 guage is typical for 10-15 ft run. Ground wires going back from the console, especially if its one wire acting as ground for many instruments, needs to be larger, and they typically run 10awg. hope this made sense, and you followed my math logic.

[jlinder]

Ron7000,We have an interesting discussion going here. I suspect we only have the real diehards left after all this math.There is an error in your math. When you totaled up the voltage drop you forgot Rw3. This will give you another 2.22v drop, making it not 10.1 v to the device, but 7.88v.You are also using 13.5v as the source voltage, taking charge voltage into account. I used the battery voltage of 12.6 plain. If you had started with 12.6 the voltage would be closer to 6.98v by your calculations. We are still differennt by about 1 volt, but we do agree that that is no way to power equipment.Another factor is heat - the wires are heating up. I think you showed about 1.3w/ft of wire using the gauge mentioned. Maybe not too much for a single wire.Now consider that you have a wire out and back, typically bundled as a pair. That makes 2.6w/ft.To that add the fact that wires are typically bundled together. Even more heat.Another problem is that your calculations assume that the device is a fixed resistance. That is not the case a lot of times. Light bulbs have a lower resistance when cold then when hot. Electronic devices have voltage regulators that work to maintain constant voltage, and they do this by drawing more current at lower voltages to keep the power constant. This makes the problem even worse. When all is said and done, I think we agree that there are probably a lot of boats out there with wiring that owners have put in that can cause real problems.

[ron7000]

sorry, there were errors in my post. I went back and editted in bold. Biggest error was I didn't add up Resistance Total right, should be 1.03 ohms not 0.964 ohms. That in turn threw everything off.I think you got confused with my 13.5 being volts. It was amps. That value is now 12.621 when using the correct resistance total of 1.03 ohms.

[bhile]

That is all well and good and what scares me is I understand it. However, the bottom line is -- How do you get a 4 gauge wire in an 8 gauge hole??

[ron7000]

Originally posted by Jack L:We have an interesting discussion going here. I suspect we only have the real diehards left after all this math. ya think? I used the battery voltage of 12.6 plain. If you had started with 12.6 the voltage would be closer to 6.98v by your calculations. I don't think so. I'd have to do the math. What caught my eye and made be deicide to post was your calculations showed about 50% drop in voltage to the device. It's less than 25% actually. The killer is all the resistance in the circuit, but there is one good thing about it- will explain below.We are still differennt by about 1 volt, but we do agree that that is no way to power equipment.CorrectAnother factor is heat - the wires are heating up. I think you showed about 1.3w/ft of wire using the gauge mentioned. Maybe not too much for a single wire.Now consider that you have a wire out and back, typically bundled as a pair. That makes 2.6w/ft.To that add the fact that wires are typically bundled together. Even more heat.Another problem is that your calculations assume that the device is a fixed resistance. That is not the case a lot of times. Light bulbs have a lower resistance when cold then when hot. Electronic devices have voltage regulators that work to maintain constant voltage, and they do this by drawing more current at lower voltages to keep the power constant. This makes the problem even worse. When all is said and done, I think we agree that there are probably a lot of boats out there with wiring that owners have put in that can cause real problems.

I don't think the problem is that bad, at least in the sense a fire is going to happen. Usually the result is equipment that doesn't work, because of a lack of power.With devices that vary in resistance, usually the resistance doesn't vary enough so voltages and currents don't change significantly. My math error shows that, although it's only a few tenths of an ohm. Again, would have to do all the math. I guess it would have to depend on the equipment being wired.What's interesting about this scenario is the 25 ft ground wire on the ground side of the device. It causes a voltage at the device to be 10.2 volts, with circuit current of 12.6 amps. If you remove that ground wire resistance (make it 0.04 ohms for example), voltage at the device drops to 9.9 volts but circuit current increases to 14.36 amps. It's here, when you start lowering the resistance that more current starts flowing through the thinner (18awg) wire and heat would become more of a problem. As for how much wattage is too much, if you find a reference table somewhere let me know. But wouldn't the heat be spread across the length of the wire? If that's the case, then longer lengths of wire can handle more power? Here I'm only talking about before the wire melts and insulation catches fire. At this point, any electronics wires on the line are definitely not gonna get enough power.

[swist]

The temperature rating for even the cheapest modern insulation is pretty high. Without looking a a piece of Ancor wire I suspect marine wire is even higher. Yes you should worry about (somewhat) undersize or over-length wires heating because that will increase their resistance even more and you may reach a point of unacceptable voltage drop, but I think you will find the main cause of meltdowns and fires is unfused shorts. Proper fusing of low-voltage wiring, boats and elsewhere, is a more recent concept than for high voltage. In my field, we use a lot of 12VDC wiring and the National Electrical Code has only relatively recently (last 10 years) talked much about low voltage wiring. A lot of old boats are underfused - either by design, or by incompetent amateur work.