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Re: Electrolytic Rust Removal

To: Ken Streeter <streeter@sanders.com>
Subject: Re: Electrolytic Rust Removal
From: "Michael D. Porter" <mdporter@rt66.com>
Date: Thu, 22 May 1997 02:31:51 -0700
Cc: triumphs@Autox.Team.Net
Organization: None whatsoever
References: <winATT-2.7.1-jrunge-56> <33833C03.1F9B@sanders.com>
Ken Streeter wrote:
 

> I wonder how much current that would take, and for how
> long?

There's a simpler formula for how much energy it takes to break an ionic
bond, but I can't find it right now. Maybe a longer way around it is
this way:

Rust is Fe2O3, with a gram/molecular weight of 160 grams. A pound is 454
grams. Therefore, one pound of rust is 2.84 moles. Since Avogadro's
number is 6.02 x 10^23 molecules per gram/mole, this would mean 1.71 x
10^24 molecules. To reduce the iron oxide to iron, an exchange of 6
electrons is required, so one needs roughly 1.03 x 10^25 electrons. An
electron has a charge of 1.6 x 10^-19 coulombs, so electrons needed
times the charge on the electron gives a total charge required of 1.65 x
10^6 coulombs. One ampere = one coulomb per second, so, if the
electrolyte were concentrated enough to, say, produce a current flow of
10 amps at 12V, it would take about 45.8 hours to reduce one pound of
rust to iron. About 6-7 kwh with conversion losses in the battery
charger; depending on your electrical rate, about a dollar.  

Then throw in about $15-20k for the swimming pool, a couple of hundred
bucks worth of potassium hydroxide, several hundred for non-magnetic
stainless steel wire mesh to make the positive electrode, and you're in
business. <g>

Cheers, Ken.  

-- 
My other Triumph doesn't run, either....

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