fusible link questions

[Tom Parker]

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. (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!

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/