Sorry, I goofed in my previous post on calculating the current density.
Here's a better calculation:
[...]
> The anode calculates out to a little over 1 square inch:
> A = pi * d * l
> = pi * 0.125 * 3
> = 1.1780972 sq. in.
That didn't sound right, so I looked up the actual wire gauge and
list it here for reference:
Gauge Diameter Resistance
(AWG) (in) (Ohms/M-ft.)
-----------------------------------
0 .3249 .09827
1 .2893 .1239
2 .2576 .1563
3 .2294 .1970
4 .2043 .2485
5 .1919 .3133
6 .1620 .3951
7 .1443 .4982
8 .1285 .6282
9 .1144 .7921
10 .1019 .9989
11 .09074 1.260
12 .08081 1.588
13 .07196 2.003
14 .06408 2.525
15 .05707 3.184
16 .05082 4.016
17 .04526 5.064
18 .04030 6.385
19 .03589 8.051
20 .03196 10.15
21 .02846 12.80
22 .02535 16.14
23 .02257 20.36
24 .02010 25.67
25 .01790 32.37
26 .01594 40.81
27 .01420 51.47
28 .01264 64.90
29 .01126 81.83
30 .01003 103.2
The diameter of my rods measure 0.80" in the unused portion, and
0.75" at the tips. They are rough and pitted after a year of use, so
I will estimate the current density using the new wire diameter and
the actual measured wetted length:
A = pi * d * l
= pi * 0.080 * 3.85
= 0.96761054 sq. in.
The current density is then:
D = I / A
= 1.15 / 0.96761054
= 1.19 mA / sq. in.
Close enough:)
Best Regards,
Mike Monett
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