WorldWide Drilling Resource

There’s a lot of confusion about wire sizing - specifically, why we use a certain size wire for a certain application. It all has to do with the cost of the wire (conductor). We want to use a conductor that will handle the amperage load adequately, but isn’t too big and costly. What do we mean by “handle the amperage load adequately”? We select conductor sizes to be big enough (wire diameter) to con- duct the amperage load with minimal voltage losses from the voltage source to where the power is being used - what is called “the load”. Typically, we want to keep our voltage losses under 3% of the source voltage. So if we have 240 volts at the source, we want to lose no more than 7.2 volts at maximum continuous amperage load. [240 (volts) x 0.03 (3%) = 7.2 volts] We want minimal voltage losses for a number of reasons. With motors, it is important to keep the voltage within 10% of the nameplate voltage. So while there will be some transmission losses, a 3% voltage reduction won’t be a problem for the motor. Another reason we want to keep the voltage losses small is because this loss represents wasted energy. This wasted energy is also dissipated as heat and, if the conductor isn’t fused correctly, the conductor could heat up - possibly becoming a fire hazard. So how do we determine the correct wire size for the application? The formula to determine voltage loss through a copper conductor is: Voltage drop through the conductor = 21.4 x Feet x Amps ÷ Circular Mils Where: 21.4 = Resistance Constant. Copper has a resistance of 10.7 ohms per foot, per circular mil at 68ºF (20ºC). This value is doubled because we have electrical resistance both to and from the load. Feet: Length of conductor from the source voltage to the load. Amps: Maximum continuous amperage the conductor is expected to carry, and is fused for. Circular Mils: 1 circular mil = The diameter of a round conductor measuring 1/1000 of an inch. A circular mil chart listing equivalent circular mil values for different AWG (American Wire Gauge) wire sizes can be found on the Internet. So basically, the wire is just a big resistor. It’s a very low value resistor, but most certainly a resistor. All materials have some level of electrical resistance. There is no such thing as zero resistance because there is no perfect conductor. How do we know what actual resistance is? The resistance of any material depends on the material type (like copper or aluminum), the material length from voltage source to load, and the material cross-sectional area. The most common measurement for conductor cross-sectional area is the circular which we discussed earlier. So to determine the resistance of 100 feet of #12 AWG copper wire/cable, we take the resistivity of this material which we know to be 21.4 ohms per foot per circular mil and divide it by the number of circular mils, which for #12 AWG cable is 6530. 21.4/6530 = 0.00327 ohms per foot. We multiply this by 100 because we have 100 feet of conductor which makes the resistance 0.327 ohms for the 100 foot length. Ohm’s law tells us that Voltage = Amperage x Resistance. So if we want to transmit 10 amps from voltage source to load, 10 x 0.327 ohms would be 3.27 volts loss over that distance. If our voltage source was 120 volts, we would be less than the 3% voltage loss allowable as 120 x .03 = 3.6 volts. Ohm’s law also tells us that Volts xAmps = Watts, so about 400 watts is being dissipated over its length. 120 x 3.27 = 392. One can also “reverse engineer” this concept and estimate amperage loads by measuring the difference in voltage from source to load. If we see a 5.0 volt loss when the load is “on” and we have 50 feet of #12AWG cable, we take the resistance per foot and multiply it by 50. So if, 0.00327 x 50 = 0.163, we have 0.163 ohms resistance over the 50 foot length of run. Ohm’s law says Voltage ÷ Resistance = Amperage. So, 5.00 ÷ 0.163 = 30.67 Amps We know this conductor is carrying about 31 amps, and it doesn’t matter what the source voltage is. All we’re interested in is the voltage difference. The (voltage at the source) - (the voltage at the load). This works equally well withAC or DC voltage. This also explains why low-voltage power generation like solar panels and wind turbines is so expensive. The conductors have to be so large to transmit even small amounts of power with acceptable energy losses most people would deem large-scale generation of this sort impractical. The bottom line with wire and cable sizing is the wire/cable needs to be large enough to carry the load with acceptable energy losses, yet no larger; because larger cable costs more and provides no additional benefit. In this case, bigger isn’t necessarily better. If we pay 100% more by going to the next larger cable size, to go from a 3% power loss to a 1½% power loss, we pay twice as much to gain about 1% less power loss. Not a very good trade-off. Britt Britt Storkson may be contacted via e-mail to michele@worldwidedrillingresource.com "0",-" - 0 &' '" &) (*-. -&2"- )! )/( ", *# ' !"- "1 -"! &.- *'" +")",- , $ &.- &)$- ((",- ((", &.- &0",.",- " .&++&)$ . &'&2",- ' 1 &.- &+" &+"- *'. *) "".% / - * &'" &. *&) ,*! %*.( &' *( 111 &. *&) /- Wire Sizing Revisited by Britt Storkson Owner, P2FlowLLC 30 OCTOBER 2015 WorldWide Drilling Resource ®