Date: Thu, 15 Jun 2000 19:19:52 EDT
Reply-To: FrankGRUN@aol.com
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: Frank Grunthaner <FrankGRUN@aol.com>
Subject: Re: 2nd batt. charging and relays. The whole truth
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Had to step in here too. Beckett is exactly correct.
From another point of view, the gauge of wire and its length determine the
resistance of the wire. The voltage dropped across the wire connected to the
battery or primary 12 V source is given by V = i x R, (classic and basic). As
more current flows across the wire, the wire heats up, the resistance goes
up, the current drops. For 15 amps drawn through a 0.5 ohm resistor, the
voltage drop would be 7.5 volts. The voltage applied to the dead battery
would be 12 V - 7.5 V or 4.5 V. The resistance in the problem is actually the
full resistance to the common ground shared with the voltage source. This
includes the resistance of the connecting wire going to the relay, the relay
contact resistance, the resistance of the connecting wire from relay to
(presumably) dead battery, the resistance of the dead battery and finally the
resistance of the battery ground terminal to common ground.
Assuming the supply voltage from the alternator is 14.4 V max, and the total r
esistive path is 1 ohm then the maximum current flow that can be supported by
this connection is 14.4 amps. (voltage drop would be the full 14.4 volts).
Typical resistivities of fully discharged deep cycle batteries are around 0.3
to 0.5 volts. If we take the value inferred by a 75 amp current draw,
assuming no resistance in the connecting wiring and an applied voltage of 12
V, then the resistivity of the dead deep cycle is/was 12 V/ 75 A = 0.16 ohms.
With this model and going back to the 14.4 V applied by alternator, the Hella
fuse is quite adequate so long as the total resistance of the auxiliary
battery circuit is 0.96 ohms. Subtracting the resistance of the battery gives
a net of 0.8 ohms. 10 gauge wire has a resistance at 0 C of about 0.01 ohms
per foot, 12 gauge wire is about 0.02 ohms per foot, 14 gauge is 0.04 and 18
gauge is about 0.1. If the distance to the battery source is approximately 10
feet, then the resistance of 18 gauge wire exceeds 1 ohm.
The resistance of the relay contacts and the connectors easily sums up to 0.3
to 0.4 ohms. 10 feet of 14 gauge wire will quench the flow of 15 amps.
Some caveats: The resistance of the battery is not a classical resistance but
more of a chemical equivalent of work done at a particular surface voltage.
But not relevant here. Remember as the battery charges, the voltage of the
battery comes up and the applied voltage over the system resistance decreases
and the current decreases.
So what is dangerous:
1. Applying more than 14.4 V+ to the electrical system. Will burn out many
expensive things.
2. Wrong polarity. Equivalent to summing the voltages (12 V battery with 12 V
applied ass-backwards is similar to 24 V applied). Smoke, damage,
destruction.
3. Connecting dead auxiliary battery to alternator or charged battery with
3/8 inch cable of stranded copper. Will support current of 75 amps plus. This
will kill VW/ Bosch alternators in no time.
4. Using number 2 screwdriver to short across fully charged lead acid battery
to see if its really charged. Gives lead plasma, sometimes with significant
atomic emission from Carbon, Nitrogen, Sodium, Phosphorus (former human body
parts). Very similar to 3 above.
To summarize, Ron's auxiliary battery kit is just fine. Use it with comfort,
follow the directions, live in peace!
Sorry, couldn't help it.
Frank Grunthaner
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