cable sizing

I need a 100 feet extension cord for my ups system and on Amazon there are 2 offerings from the same manufacturer – 100 feet of 12/3 for $59 or 100 feet of 14/3 for$45 both are available via Amazon Prime, so there is no shipping cost.  What is the difference and which should I buy?

I know there are cable sizing online calculators, so I entered my requirements – a 30 meter length (98 feet), a current of 3 ohms, 12 volts dc, and a drop in voltage or power loss not to exceed 4%.  And I got very different answers.  Two sites recommended I use 8AWG and one site 12 AWG.  8AWG (8.35mm^2) is about 2.5 times thicker (and probably at least twice as expensive) as 12AWG (3.3mm^2), which was recommended by a 3rd site.  A conundrum – what to do?

First, terminology – 12/3 is 12 gauge cable containing 3 wires and 14/3 is 14 gauge cable containing 3 wires.  Using metric, because it is much easier for calculations, 12 gauge has a cross sectional area of 3.31 mm^2 (that’s millimeters squared) and 14 gauge is 2.08.  So the 12 gauge has almost 60% more copper than the 14 gauge  $\frac { 3310-2080 }{ 2080 } =59.1%$.  But the cost is only 31% more and this could be because the product consists of other components such as the plug and socket which are the same cost for both, and cable covering which may increase only slightly.

When grid power is out, my ups battery powers the components and so I want to minimize unnecessary loss of power in order to extend the life of the battery during the outage.  A source of loss is using undersized cable to carry the current.  If wires are undersized they heat up – think of the filament in a toaster or in the old light bulbs – so the current in the cable is transformed by the resistance of the wires into heat.

As I know too well from my gravity fed irrigation system, I lose water flow/pressure with smaller diameter pipes and with longer pipes.  Therefore the two determining factors are the cross area of the pipe (A) and the length of the pipe (L).  It is the same with electricity.  Resistance to the flow of electricity Rcable increases as L increases and decreases as A increases.  In fact Rcable doubles if L doubles and halves if A doubles.

For dc (direct current as from a battery) there are 2 useful formula from my high school days:

V = I*R     (voltage = current (amps) * resistance (ohms).

P = V*I     (power (watts) = voltage * current (amps)).

The power lost by a cable can be calculated as follows:
${ P }_{ cable }=\triangle { V }_{ cable\\ }\ast I\\ \triangle { V }_{ cable }=\frac { { R }_{ cable } }{ { R }_{ cable\quad }+{ R }_{ load\\ \\ } } \\ since\quad V=I\left( { R }_{ load }+{ R }_{ cable } \right) \\ therefore\quad { P }_{ cable }={ I }^{ 2 }\ast { R }_{ cable }$

So if the current carried by a cable doubles – say from 2amps to 4 amps the  cable resistive loss will increase by 4 times i.e. $\frac { { 4 }^{ 2 } }{ { 2 }^{ 2 } } =\frac { 16 }{ 4 } =4$

(for principles thanks to edX MOOC given by Delft University of Technology on Solar Energy)

To calculate cable resistive loss I need cable length, cable cross area, current carried and the specific resistance or conductivity of the cable.  Conductivity is the inverse of specific resistance which means as the one increases, the other decreases – the greater the conductivity the lower the resistance.  The specific resistance of copper is 0.0168 ohms/mm^2.

Assume 12 gauge copper cable is 30 meters long and is carrying 36 watts power (3 amps at 12vdc).

The cable resistance calculation is straightforward – specific resistance*cable length/cable area – 0.0168*30m/3.31mm^2= 0.152.  The cable power drop is I^2*Rcable = 9*0.152=1.37 watts. So the power loss is 1.37/36 = 3.81%. Another way to calculate the power loss % is voltage drop/total voltage =(0.152*3)/12=3.8%.

If instead of using 12 gauge cable I decided to save money and purchase 14 gauge the only change to the calculation is to substitute 2.08 for 3.31 representing the cross sectional area.  The cable resistance would be 0.0168*30m/2.08mm^2= 0.242, the cable power drop is 9*0.242=2.178 watts and the power loss is 2.18/36 =6.1%.  So an extra 2.3% power loss from using the next size smaller gauge wire.

Now these are just my calcs and I am not qualified in this area, but as mentioned at the beginning the calculations of one website came in very close to my own.  So how can another site recommend 8 AWG.  I contacted the overseas site and received a very prompt courteous reply:

“Ah, we have had this question a few times.

There are variable factors when calculating the voltage drop of a cable, such as whether it is held in free air, conduit or buried in walls, what the ambient temperature is, etc.  Our drop calculation uses our selected default values for these factors.  That may well be the difference, you or your formula are using different values for these factors.

Our cable sizing is a guideline figure, there are sites that allow you to get more specific with the above factors, setting them yourself to reflect your true situation.”

My suggestion is find a reputable website and enter your requirements and go with their recommendations.  And I welcome any comments on my calculations above.