Re: 42V Car Batteries

"Spehro Pefhany" <speffSNIP@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:q9mmi3lto6f4darg4504c4cv1171lnp6l5@xxxxxxxxxx
I guess you'd have to work out where most of the voltage drop would
occur to see where the power is going. Anyone care to measure the
resistance of, say, a 12" adjustable wrench?

I would be more concerned about the contact resistance, as well as what the
terminals are made of. Lead isn't that great a conductor, and the junk
they toss in doesn't make it any better a conductor nor keep it from
melting.

Offhand, the resistivity of iron is about 100 nohm.m, so for a 12" bar of
about 1/4 x 1/2" = 1/8 in^2 dimension, you get 0.8 cm^2, or 8 x 10^-6 m^2
cross section, and about 0.3m length. 100 nohm.m * 0.3 m / 8 x 10^-6 m^2 =
3.72 x 10^6 nohm = 3.72 mohm.

12V / 0.00372 ohm = 3225 amperes, so contact and internal resistance are
going to be a big factor, if not the limiting factor.

On the upside, if contact resistance is good, much of the power will in
fact go into heating the battery itself.

On the downside, iron's resistivity increases rapidly with temperature.
Once the iron hits 100C, resistance is up by 60%, and power is up by 60% or
more.

Iron's specific heat capacity is 0.45 J/g.K, or 450 J/kg.K. The 12 x 0.25
x 0.5" bar weighs about 4.5 pounds or 2 kilos, so the energy for that 100C
temp rise is 90kJ. If the short-circuit current is 500A, the power
dissipation is 930W, which will heat up the bar by 100C in somewhat less
than 100 seconds, since the resistivity and power dissipation increase with
temp rise. Well, that's good news for the wrench-holder, ignoring contact
resistance anyway. In the mean time, the other 10V dropped across the
battery will have heated it a little bit, but considering the weight and
water content, I couldn't say if it's enough. But it doesn't need to get
as hot, either.

Tim

--
Deep Fryer: A very philosophical monk.
Website @ http://webpages.charter.net/dawill/tmoranwms


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