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Wire-Gauge Ampacity

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1 Wire Gauge - Ampacities

[edit] Wire Gauge - Ampacities

[edit] AWG

American Wire Gauge (AWG) Diameter where:

$d =$ diameter in Inches
$n =$ AWG gauge number
$A =$ cross-sectional area in sq inches

$d = \left( \frac{460}{100} \right)^{ \frac{ 36 - n }{39 } }$
$n = -10.7 + 10\log_{10}{\frac{1}{A}}$
$A = 10^{-\frac{n+10.7}{10} }$

[edit] Load Carrying Capacities or ampacities

In part from Reference Data for Engineers: Radio, Electronics, Computer and Communications 7th Ed

Max ambient 60C

AWG	Dia Inch	Cir Mil	Dia cm	Area Inch²	lb/kft	ohms /kft	Ohms /km	CU Max free-air Amps	CU Max enclosed Amps
32	0.008	63.2	0.020	4.964E-05	0.19	199.587	654.613	.53	0.32
30	0.010	100.5	0.025	7.894E-05	0.30	125.521	411.690	.86	0.52
28	0.013	159.8	0.032	1.255E-04	0.48	78.941	258.915	1.4	0.83
26	0.016	254.1	0.040	1.996E-04	0.77	49.647	162.833	2.2	1.3
24	0.020	404.0	0.051	3.173E-04	1.22	31.223	102.407	3.5	2.1
22	0.025	642.4	0.064	5.046E-04	1.94	19.636	64.404	7.0	5.0
20	0.032	1,021.5	0.081	8.023E-04	3.09	12.349	40.504	11.0	7.5
18	0.040	1,624.3	0.102	1.276E-03	4.92	7.767	25.473	16	10
16	0.051	2,582.7	0.129	2.028E-03	7.82	4.884	16.020	22	13
14	0.064	4,106.7	0.163	3.225E-03	12.43	3.072	10.075	32	17
12	0.081	6,529.9	0.205	5.129E-03	19.77	1.932	6.336	41	23
10	0.102	10,383.0	0.259	8.155E-03	31.43	1.215	3.985	55	33
8	0.128	16,509.7	0.326	1.297E-02	49.98	0.764	2.506	73	46
6	0.162	26,251.4	0.412	2.062E-02	79.46	0.481	1.576	101	60
4	0.204	41,741.3	0.519	3.278E-02	126.35	0.302	0.991	135	80
2	0.258	66,371.3	0.654	5.213E-02	200.91	0.190	0.623	181	100
1	0.289	83,692.7	0.735	6.573E-02	253.34	0.151	0.494	211	125
0	0.325	105,534.5	0.825	8.289E-02	319.46	0.120	0.392	245	150
00	0.365	133,076.5	0.927	1.045E-01	402.83	0.095	0.311	283	175
000	0.410	167,806.4	1.040	1.318E-01	507.96	0.075	0.247	328	200
0000	0.460	211,600.0	1.168	1.662E-01	640.53	0.060	0.196	380	225

[edit] House Wiring

First, avoid the most common error in house wiring; that is hanging a 120Vac outlet off of a 3 conductor (L1,L2,N) 220Vac cable - it is dangerous! To do something like that you need a 4 conductor cable - (L1,L2,N,G) toprovide a safety ground that is _NOT_ carrying any current.

The table below will not support any reduced ground cables and is for _Copper_wire_only_!. If you run aluminum wire you need to consult the manufactures data sheets and be sure to understand terminal block treatments! This table also in no way compensates for length of run! This table is not for wires packed in conduit - wires need to dissipate heat!

Normal House Wiring Gauge Consult your building inspector because your location will have a specific building code!
Circuit Maximum Amperage	Minimum COPPER wire gage
15A	14AWG
20A	12AWG
30A	10AWG
45A	8AWG
60A	6AWG
80A	4AWG
100A	2AWG
If you are running more than 100A you probably will run Aluminum and need a different table. Do not use copper ratings for aluminum!

[edit] PCB Trace Width vs Current

[edit] Fusing Point of Wire

H. W. Preece’s Investigation way back in 1884 provided a way to calculate the fusing current of a wire based on its diameter.

$I = kd^{\frac {3}{2}}$

$d = \left ( \frac{I}{k} \right ) ^{\frac{2}{3}}$

Where:

I is the current in Amps

d is the diameter in cm

k is the fusing constant

Material	k for d in cm	Melting point°C	Resistivity (Ω-m) at 20 °C	Coefficient*
Silver	1900	961.78	1.59×10^-8	.0038
Copper	2530	1084.62	1.72×10^-8	.0039
Aluminum	1870	660.32	2.82×10^-8	.0039
Iron	777.4	1538	1.0×10^-7	.005
Tin	405.5	231.93	1.09×10^-7	.0045
Platinum	1277	1768.3	1.1×10^-7	.00392
Lead	340.5	327.46	2.2×10^-7	.0039
Tin/lead solder	325.5	183	1.44x10^-7	.0035

There should be a relationship between resistivity, melting temperature and k. Remember that the electrical resistivity ρ (Rho (letter)) of a material is given by:

${\rho={R \left. \frac{A}{\ell} \right.}}$

where

ρ is the static resistivity (measured in ohm meters, Ω-m);

R is the electrical resistance of a uniform specimen of the material (measured in ohms, Ω);

$\ell$ is the length of the piece of material (measured in meters, m);

A is the cross-sectional area of the specimen (measured in square meters, m²).

[edit] Watts generated in a Wire

First, an example using the formula in the above section to determine the wattage dissipated in a 1 meter length of 12 gage copper wire at 10A.

${\rho={R \left. \frac{A}{\ell} \right.}}$

Rearrange to solve for R

$R = \frac{\rho\ell}{A}$

Remembering that $P = I 2 R$

$P = \frac{I^2 \rho\ell}{A}$

Substituting in where:

I = 10

ρ = 1.72 x 10 - 8 Ω - m

$\ell = 1m$

d = 0.205 c m = .00205 m

$r = .001025m \approx .001m$

$A = \pi r^2 \approx 3.14 \mathrm{x} 10^{-6}m^2$

$P \approx \frac{\left (10^2\right) 1.72\mathrm{x}10^{-8} \left ( 1 \right)}{3.14 \mathrm{x} 10^{-6}} \approx 0.55W$

[edit] Heat Rise in Wire in Still Air for the same wire

Now, to figure the temperature rise per watt.

$\dot{Q}_{12} = \epsilon A\left ( \sigma T_1^4 - \sigma T_2^4\right )$

Where :

$\dot{Q}_{12} =$ Net radiant energy

ε =

emissivity (sometimes called emittance) - the constance of likeness to a black body

ε = 1

a blackbody

ε = 0.35

a very shinny aluminum alloy

T =

Absolute temperature in K

$\sigma = 5.67 \mathrm{x}10^{-8} \frac{W}{M^2 K^4}$ Stefan-Boltzmann constant

When $T 1$ and $T 2$ are not very different, it is convenient to linearize the equation by factoring the term $\left( \sigma T_1^4 - \sigma T_2^4 \right )$ to obtain:

$\dot{Q}_{12} = \epsilon A \sigma \left( T_1^2 + T_2^2 \right )\left(T_1 + T_2\right )\left(T_1 - T_2\right )$

$\dot{Q}_{12} \approx \epsilon A \sigma \left( 4T_m^3\right)\left( T_1 - T_2 \right )$

When $T_1 \approx T_2$ and $T m$ is the mean of $T 1$ and $T 2$

Written again as

$\dot{Q}_{12} \approx A h_r \left( T_1 - T_2 \right)$

Where

$h_r = 4 \epsilon \sigma T^3_m$ or Radiation heat transfer coefficient with the units $\frac{W}{m^2K}$

$h_r = 4 \epsilon \left(5.67 \mathrm{x} 10^{-8} \frac{W}{m^2K^4}\right)\left(298K\right)^3$

$h_r \approx 6 \epsilon \frac{W}{m^2K}$

Emissivity is always a fuzzy number that depends on details like surface finish and for copper can vary from 0.7 to .88 but for electrical work 0.4 works most of the time. Give it large error bands!

The surface area of our wire is

$A = \ell \pi d = \left(1m\right) .0205m \pi = .0644 m^2$

Our power is 0.55 Watts and $\epsilon \approx 0.4$

$T_2 = 30^\circ C = 303K$

$\dot{Q}_{12} = \epsilon A\left ( \sigma T_1^4 - \sigma T_2^4\right )$

$0.55W = \left ( 0.4 \right ) \left ( 0.644m^2 \right ) \left [ \sigma T_1^4 - \left ( 5.67 \mathrm{x} 10^-8 \frac{W}{m^2 K^4} \right ) \left(303K\right)^4 \right ]$