fusible link questions

[Tom Parker]

Clarence L.Snyder <clare.snyder.on.ca at ibm.net> wrote:

>> 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.

What is the current in and voltage across the field windings necessary to
generate the power proposed? I'm willing to bet that it isn't that huge.
Certainly the full power output proposed by the original poster will not be
dissapated in the field windings.

I've no idea what current is normally presant in the field windings, but I
wouldn't expect it to be large, and I would expect that it could be increased
significantly before you have problems with overheating.

The other issue noone has mentioned is the voltage output, can the diodes and
other components survive the high voltage?

--
Tom Parker - tparker at nznet.gen.nz
           - http://www.geocities.com/MotorCity/Track/8381/