fusible link questions

Clarence L.Snyder clare.snyder.on.ca at ibm.net
Sun Feb 7 02:13:28 GMT 1999


Tom Parker wrote:
> 
> Clarence L.Snyder <clare.snyder.on.ca at ibm.net> wrote:
> 
> >> A switching regulater, when reducing voltage, will increase the current
> >> (much like a transformer).  My idea is to max out the field coil on a
> >> standard alternater, which will produce a high output voltage (> 100 V).
> >> The alternater can still supply its rated current at the higher voltage
> >> because the current is limited mostly by I2R losses.
> 
> >Actually, the alternator is rated for maximum POWER output. 100 amps at
> >14 volts is 1400 watts, almost 2 HP. At an optimistic 80% efficiency, it
> >requires closer to 2.5 HP. Pushing the same current at higher voltage
> >quickly oveheats the alternator due to the increase in these I^2 R
> >losses, which increase at the rate of a power of two with voltage
> >increase - double the voltage, 4 times the heat with the same current.
   
> Erm... no.
> 
> You've just said that the losses due to I^2 R depend on voltage, when there is
> no voltage term in your equation.
> 
> What does matter is the current handling ability of the field windings. The
> original poster proposes to increase the current in the field windings while
> using some devious switching transformer to keep the output current of the
> main windings down, and also to step the voltage down to something that won't
> fry the electronics in the rest of the car.
> 
> The only extra electrical load on the alternator will be in the extra current
> flowing in the field windings. The field windings will have to dissapate heat
> due to the P = I^2 R equation, so if you double your field winding current,
> you will have to dissapate 4 times the heat in them.

Mabee I didn't explain it 100%, but I THINK that's what I said. Double
the power output of the alternator = 4 X the heat to dissipate, due to
I^2 R losses.
 (there are also extra
> mechanical loads, you will stress the bearings and the shaft of the alternator
> if you ask it to draw such a huge amount of power out of the engine).
> 
> If you double the voltage output of the alternator main windings, the heat
> dissapated in the main windings will remain the same, as the heat generated
> there is due to the current through them only!

Correct - stator current will remain constant, so I^2 R losses will
remain constant. Stator heating will not be the problem. - 
However, if more POWER is required at the same voltage, then both field
and stator heating increase at the same "square of increase" rate.

> 
> I can explain the mistake you made applying the P = I^2 R equation if you
> want, but it might start a huge discussion, which people on the mini-list will
> know about!
> 
> --
> Tom Parker - tparker at nznet.gen.nz
>            - http://www.geocities.com/MotorCity/Track/8381/



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