
ContinuousWave Whaler Moderated Discussion Areas ContinuousWave: Small Boat Electrical Wire Conductor Size for Power Distribution

Author  Topic: Wire Conductor Size for Power Distribution 
jimh 
posted 05112013 01:00 PM ET (US)
Wire Conductor Size for Power Distribution In distribution of power, there are two concerns in choosing the size of the conductors that will carry the electrical current: the rating of the conductor for current capacity the voltage drop that will occur in the conductor In systems where the distributed voltage is low, the concern for the voltage drop tends to limit the length of the conductor that can be used. This article examines this relationship in more detail. The resistance of a wire for a certain unit of length is a function of the material used for the wire and its cross section area. For a conductor made of copper and sized according to the American Wire Gauge (AWG standard) we can find the resistance per 1,000feet of conductor from a table. See http://www.powerstream.com/Wire_Size.htm for a good listing of this data. If we use a wire of 10AWG, we see that 1,000feet of 10AWG wire will have a resistance of 1Ohm. This value is really a physical property of copper, so there is not much variance seen in listings of the resistance of copper wire. The current carrying capacity or current rating is a value that is more difficult to define with precision. As current flows through a conductor it encounters resistance, which results in a power loss. The power is converted to heat. The wire becomes warmer. The amount of temperature increase of the wire above the ambient temperature becomes a factor in rating the maximum current. If the conductor is a single conductor and is in free air space, much more heating can be tolerated. If the conductor is in a bundle with other conductors and enclosed in a conduit with little cooling air, the maximum current allowed will be lower. The type of insulation also affects the current rating. Expensive insulation material like Teflon can tolerate higher temperatures. The result of the many variables affecting wire temperature is the maximum current rating is more variable. For our example conductor of 10AWG, we see the maximum current for a single conductor in free air is 55Amperes. For other situations, such as power distribution, the maximum current is derated to only 15Amperes. When a current flows in a wire, the voltage drop in the conductor will be equal to the resistance of the wire in Ohms times the current in the wire in Amperes. For wire of a particular gauge, we can compute the resistance by the length of the wire and the resistance of that wire gauge per 1,000feet. We can say (1) VoltageDrop = LengthinFeet x (Resistance/1,000feet) x Current Typically in power distribution a limit on the voltage drop to a certain percentage of the system voltage is required. A limit of threepercent is often used. We can set this as our maximum allowed VoltageDrop by VoltageDrop = SystemVoltage x 0.03 In the case of a 12Volt system, the result is VoltageDrop = 0.36Volts Now we can calculate the LengthinFeet of a particular conductor that will create a voltage drop of 0.36. We solve equation (1) for LengthinFeet: (2) LengthinFeet = VoltageDrop / [(Resistance/1,000feet) x Current] This rearranges to (3) LengthinFeet = VoltageDrop x 1000 / (Resistance x Current) Now we explore a 10AWG conductor at its maximum rated current of 55Amperes in a 12Volt system LengthinFeet = 0.36Volt / [1Ohm/1,000feet) x 55Ampere] LengthinFeet = 6.54feet Because this is the total length of the conductors, in a twoway circuit, the maximum distance that a 10AWG conductor could carry its rated 55Amperes before the voltage drop in a 12Volt system exceeded 3percent is only 3.25feet. Now we explore a current of only 15Amperes, which is the rated current for 10AWG when used in power distribution (according to the table mentioned above). (See the footnotes in the table for the criteria used to obtain this rating.) Recalculating the LengthinFeet for the 10AWG conductor before we reach the maximum voltage drop of 3percent in a 12Volt system, we have LengthinFeet = (0.36Volt x 1000) / (15Ampere x 1Ohm) = 24feet Again, for a twoway circuit, this value is divided in two, and we see that a 10AWG conductor can be used to distribute 15Amperes in a 12Volt system for only a maximum distance of 12feet. Let's examine a typical situation in a small boat with power distribution from a battery at the stern to a secondary power distribution center at the helm, a distance of about 15feet. A conductor of 8AWG is used. The resistance per 1,000feet for 8AWG is 0.6282Ohms. The rated maximum current in power distribution is 24Amperes. We find the maximum length from LengthinFeet = (0.36Volt x 1000feet/ (24Ampere x 0.6282Ohm) = 23.9feet For a twoway circuit, the maximum length is half that or about 12feet. Let's try a 4AWG conductor: the resistance per 1,000feet is 0.2485 and the maximum current is 60Amperes. The maximum length for 3percent drop at 12Volts is LengthinFeet = (0.36Volt x 1000feet/ (60Ampere x 0.0.2485Ohm) = 24.1feet For a twoway circuit, the maximum length is half that or about 12feet. I see a nice rule of thumb developing here. The maximum length of a conductor that can be used for twoway power distribution is equal to the system voltage if the current is to be at the maxim rated current of that conductor for power distribution. In other words, in a 12Volt system of power distribution, the maximum distance for power distribution at the rated current is 12feet. Let's test this new rule. We'll use a system voltage of 24Volts. A 3percent drop at 24Volts is 0.72Volts. Resolving the 4AWG case above, finds the length to be LengthinFeet = (0.72Volt x 1000feet/ (60Ampere x 0.0.2485Ohm) = 48feet For a twoway circuit, the maximum length is half that or about 24feet. Perfect! The rule of thumb works. For a system voltage of 12Volts, and for any distance longer than 12feet, you must use a larger wire size and derate the current. We can rework the relationship in equation(1) to solve for resistance per 1,000feet: (3) Resistanceper1000feet = (VoltageDrop x 1000)/(Feet x Amperes) If we want to distribute 12Volt power with a 3percent voltage drop at a distance of 40feet with a current of 30amperes, the Resistanceper1000feet must be equal or less than Resistanceper1000feet = (0.36 x 1000)/(40 x 30) = 0.3Ohm We go to the table and find the conductor size must be 4AWG (0.2485Ohm) to meet this requirement. The 40feet we calculated is twice the twoway distance, so the maximum twoway distance will be 20feet. In other words, if 12Volt power is to be distributed a twoway distance of 20feet and a current capacity of 30Amperes is desired, the wire used must be 4AWG in order to maintain the voltage drop at not more than 3percent.

jimh 
posted 05112013 01:07 PM ET (US)
I have never seen the rule of thumb I proposed above mentioned before. Therefore, I am calling it "Hebert's Rule." The rule says: In a power distribution system where a 3percent drop is the maximum tolerated, the maximum distance in feet is equal to the system voltage in Volts at the conductor's rated current capacity for power distribution. The corollary rule: in 12Volt power distribution, the maximum length is 12feet at the conductor's rated current. This rule is based on the current rating of the wire being calculated on the basis of 1Ampere per 700 circular mills of copper conductor. Another good table of wire data is 
Jerry Townsend 
posted 05192013 05:50 PM ET (US)
"Hebert's Rule" has to also address the wire gauge. [Which it doesjimh] The 3 percent rule "target" is a good guide  but just let the user make the calculation as has been done for many years. And providing the tools to make the calculation in your website would suffice.  Jerry/Idaho 
jimh 
posted 05192013 07:40 PM ET (US)
"Hebert's Rule" does address the wire gauge. It does it by the rating of the wire gauge for carrying current as a power distribution ocnductor. The rule works for any wire gauge and any system voltage. Jerrytake another look at the rule. It is really quite an amazing rule of thumb. The purpose of the rule is to eliminate the tedium of making calculations. 
jimh 
posted 05202013 10:02 AM ET (US)
Perhaps I need to demonstrate Hebert's Rule with another example. Let's investigate with a 6AWG wire. Wire of 6AWG has a resistance of 0.3951Ohms/1,000Feet, and it is rated to distribute a current of 37Amperes. According to Hebert's Rule, the length (in feet) of this conductor (twoway circuit) that will produce a threepercent voltage drop will be equal to the system voltage. For this example we use a system voltage of 48Volts. According to Hebert's Rule the maximum length of 6AWG that can be used to distribute 37Amperes with less than threepercent voltage drop will be a twoway distance of 48feet. A twoway circuit of 48feet will have 96feet of wire. The resistance of 96feet of 6AWG conductor will be (0.3951Ohms/1000feet) x 96feet = 0.0379296Ohm If a current of 37Amperes flows, the voltage drop will be 0.0379296Ohm x 37Ampere = 1.40Volts A threepercent drop in a 48Volt system will be 48 x 0.03 = 1.44Volts which gives us a margin of 0.04Volts, or precisely what we required. Hebert's Rule has successfully predicted the maximum length that a 6AWG conductor can be used to distribute 48Volts at a current of 37Ampere: 48feet (i.e., the system voltage). Let me repeat Hebert's Rule: The maximum length (in feet) of a conductor that can be used for twoway power distribution is equal to the system voltage (in Volts) if the current is to be at the maxim rated current of that conductor for power distribution as figured according to the 1Ampere/700circular mils rating system.

jimh 
posted 05202013 10:16 AM ET (US)
Applying Hebert's Rule to boat electrical power with a system voltage of 12Volts, we see that the maximum distance that power can be distributed using a conductor of a particular AWG rating will be 12feet if the circuit is to carry the current for that AWG rating according to the 1Ampere/700circular mils calculation. Because the voltage drop is a linear function, we can quickly approximate what conductor size we need to use if the distance is longer than 12feet by looking for the current rating of the next larger gauges and choosing a wire by current rating that is proportional to the distance desired for runs longer than 12feet. For example, if 12Volt power is to be distributed over a distance of 24feet, we have a ratio of distance of 2:1. We must select the conductor size in the same proportion. If the current to be distributed is to be 15Amperes, we look at the wire table and find the conductor rated for 30Amperes. I demonstrate with another example: To distribute 12Volts at a distance of 24feet with a current of 15Amperes, we must use a conductor rated a 30amperes. Looking at the wire current table, we see that 7AWG is rated for 30Amperes. Now we check our rule: The resistance of 7AWG is 0.4982Ohm/1000feet. An outandback distribution of 24feet will mean 48feet of wire. The resistance will be 48feet x 0.4982Ohm/1000feet = 0.0239136Ohm When a current of 15Amperes flow in this resistance, the voltage drop will be 0.0239136 x 15 = 0.36Volt In a 12Volt system, a threepercent drop is 12 x 0.03 = 0.36Volt Hebert's Rule and its quick length estimation corollary have precisely predicted the proper wire size to use for maintaining the voltage drop at a threepercent tolerance. 
jimh 
posted 05202013 10:24 AM ET (US)
The rated current carrying capacity of conductors for power distribution according to the 1Ampere/700circular mills calculation is as follows 00AWG = 190Amperes In a 12Volt system, the maximum distance those conductors can carry their rated current is 12feet (or 24feet of total conductor length) if less than a threepercent voltage drop is desired. 
jimh 
posted 05202013 01:01 PM ET (US)
You can also easily estimate the distance for a higher voltage drop by multiplying the 12foot distance (that is, the system voltage) by the proportional increase in voltage drop compared to 3percent. For example, if a 10percent voltage drop is allowed, we can multiply the distance by a factor of 10/3 = 3.3times. That means for the purpose of distributing current with a tolerance for a 10percent voltage drop, the conductor can be up to 12 x 3.3 = 40feet long for the rated current. As an example, if a electrical windlass at the boat can tolerate a 10percent voltage drop, and the motor typically draws 90Amperes, we can use 2AWG conductors up to a distance of 40feet. Let's check: Conductor of 2AWG has a resistance of 0.1563Ohms/1000feet. For 80feet of conductor the resistance will be 80 X 0.1563Ohms/1000feet = 0.012504Ohms With a current of 90Amperes, the voltage drop will be 90 x 0.012504 = 1.125Volts In a 12Volt system, a 10percent voltage drop is 1.2Volts; we are below that and meet the requirement of the design. 
jimh 
posted 09122015 09:21 AM ET (US)
The above material is now presented in a single article in the REFERENCE Section at Wire Conductor Size for Power Distribution 
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