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I want to implement below formula used in this Wire Gauge Calculator in an Excel sheet. Can someone show an example?

Example: system voltage = 12 V, system current = 8 A, length = 50 feet.

resistance = resistivity * length / area

If we consider 26 gauge wire, diameter of wire = 0.0159 inch = 0.001325 feet

radius = 0.01325/2 = 0.006625

a = π * radius2 = 0.000137090

resistivity = 1.68 * 10-8

Therefore wire resistance = 0.0060 Ω

The data I get is not matching with the value in the table mentioned. Can you point out what mistake I am making?

voltage drop

I am not saying my answer is correct . If you check the online calculator I posted it gives the right choice of wire gauge.

I am trying to implement the same thing in an Excel sheet.

I could use an example to explain choosing the correct value and getting an answer similar to the online value.

My intension is to calculate:

  1. resistance per length in ohm
  2. voltage drop
  3. % voltage drop
  4. % loss
  5. select area which is good for the application
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  • \$\begingroup\$ Units are a mess, and you've lost a decimal place in the 'feet radius'. Apart from that, it's ok! \$\endgroup\$
    – Chu
    Commented Aug 24, 2015 at 9:42
  • \$\begingroup\$ 1.68*10^-8 This seems like a metric resistivity. \$\endgroup\$ Commented Oct 23, 2015 at 20:30

2 Answers 2

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26 AWG is 40 ohms / 1000 foot as per this table: -

enter image description here

A 50 foot length of this wire (2 ohms) with 8A will dissipate 128 watts i.e. it doesn't sound a sensible choice. The volt drop at 8A will be 16 volts.

As for your formula I can't tell whether it agrees or disagrees with whatever reference you have.

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Ultimately you are combining metric resistivity(1.68*10^-8[\$ \Omega \cdot m\$]), the wrong circular area (a = pi* radius^2) and english units.

From AWG Tables: #26 AWG has 40.81Ω / 1000ft @ 20°C and a diameter of 0.01594 inches.

The AWG is based on Circular Mils (CM) and a modified area calculation (\$A = d^2\$) to eliminate \$\pi\$.

$$1.72 \times 10^{-8} \Omega \cdot m = 10.37 Ω\cdot CM/ft\ @\ 20 ^\circ C$$ $$d = 0.01594 in = 0.01594 in × 1000 mils / in = 15.94 mils$$ $$A = d^2 = (15.94 mils)^2 = 254.08 CM$$ $$R = {ρ ℓ \over A} = {10.37Ω•CM/ft * 1000 ft \over 254.08 CM} = 40.81 @ 20°C $$

This agrees with the given values from the AWG table. This is essentially the math you need to go forward.

$$R = {ρ ℓ \over A} = {10.37Ω•CM/ft * 50 ft \over 254.08 CM} = 2.04 @ 20°C $$

$$V_{DROP} = I R = 6A * 2.04Ω = 12.2V$$

All of the potential of the 12V source is consumed by the wire. #26 is too small.

If you want to do a spreadsheet. Decide percentage lost to wire. For Example: 5%.

$$V_{Feeder} = 5\% \ of\ V_{Source} = 5\% × 12V = 0.6V$$

$$R_{Feeder} = {V_{Feeder} \over I_{Load}} = {0.6V \over 6A} = 0.1Ω$$

$$R = {ρ ℓ \over A}$$ Rearrange formula. $$A = {ρ ℓ \over R_{Feeder}} = {10.37Ω•CM/ft * 50 ft \color {red}{* 2}\over 0.1Ω } = 10,370 CM $$

The *2 comes from you have to get to load and back. Two conductors for DC.

Do a look up table to select a wire area > 10,370 CM. Same for actual area or Ω / 1000 ft. Calculate actual resistance. The rest should be easy to figure out.

  • Wire size: #10 AWG
  • Area = 10,381 CM
  • \$R_{Feeder} = 0.099 89\Omega \$ for 100ft
  • Maximum Current: 55 A

10,381CM is too close to design area of 10,370CM, so I'd go up to next size, which is #8 (#9 is not readily available to consumers).

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