Re: Wiring, wire size, and current capacity
Originally posted by Jack L:<br /> 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.<br /><br />But with all the strict safety precautions I was completly suprised when I found out the specs for wire gauge and current capacity.<br /><br />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. <br /><br />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.<br /><br />He comes the math. I looked up the standard resistance for 18ga copper wire and got 6.6 ohms / 1000 feet...<br /><br />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).<br /><br />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?)<br /><br />Voltage drop = resistance x current.<br />Voltage drop = .3 x 20 amps<br />Voltage drop = 6.6 volts<br /><br />Voltage to device = 12.6 volts from battery - 6.6 volt drop<br />Voltage to device = 6 volts <br /><br />Wattage dissapated in wire = 6.6 volts x 20 amps<br />Wattage 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.<br /><br />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.<br /><br />Now go back into the wiring diagram of your circuit, with wire resistance at 6.6ohms per 1000 ft.<br /><br />{13.0V} + ---- [Rw1 (15ft)] -- [Rbus] --- [Rw2 (10ft)] --- [Rdevice = 0.65ohms] ---- [Rw3 (25ft)]-- Ground.<br /><br />Rw1 = resistance of 0.099 ohms of 15 ft wire<br />Rbus = guesstimated resistance of bus at 0.05 ohms?<br />Rw2 = 0.066 ohms for 10ft wire<br />Rdevice = must be 0.65 ohms to pull 20 amps at 13 volts.<br />Rw3 = 0.165 ohms for 25ft wire from device back to ground.<br /><br />Resistance Total in circuit is all R's added together = 0.964 ohms
Correction => 1.03 ohms<br /><br />In this series circuit, current is constant from + post of battery to - post of battery, using V = IR, I = 13.0 / 1.03 = 12.621365 Amps<br /><br />you 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:<br /><br />V Rw1(15ft) = 12.621 amps * .099 ohms = 1.2495 V<br />V Rbus = 12.621 amps * .05 ohms= 0.63107 V<br />V Rw2(10ft) = 12.621 amps * .066 ohms = 0.8330 V<br /><br />Voltage at device = 13.0 - 1.2495{Rw1} - 0.63107{Rbus} - 0.8330{Rw2} = 10.2864 V<br /><br />
voltage drop across device is V = I*R = 12.621 amps * 0.65ohms = 8.20388 volts<br /><br />voltage 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!<br /><br />V drop of Rw3 = 12.621 amps * 0.165 ohms = 2.0825 Volts<br /><br />subtraction 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:<br />13.0 - 1.2495 - 0.63107 - 0.8330 - 8.20388 - 2.0825 = 0.00005 volts close enough <br /><br /><br />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.<br /><br />Wattage Rw1 = 15.77, over 15ft of wire remember so heat isn't localized;<br />Wattage Rw2 = 10.51<br />Wattage 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.<br /><br />Formuals for power P (watts) is<br /> P = V * I<br /> P = I*I*R<br /> P = (V * V) / R<br /><br />for 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.<br /><br />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.