Seems we are having basic math problems again. Look at it this way (which
is presumably what Jim Pentagrid meant). You have a power rating for a
motor. The question was, how much of the *motor's power rating* is
available if it is run as a generator?

Assume the motor is 80% efficient and that it is exactly as capable of being
a generator as it was a motor; i.e., also 80%. Take the input power rating
for the motor X 80%. This is the input power rating of that motor ran as a
generator. Incidentally that amount of power is input as mechanical motion.
Now, the motor-as-generator, is 80% efficient. The mechanical input power
is converted to electrical power at 80%. That would be 80 X 80 or aprox.
64% of the original input power rating for the motor.

Bob Swinney



"Gary Coffman" wrote in message
...
On Tue, 26 Aug 2003 22:57:37 GMT, "Bob Swinney"
wrote:
Gary sez:

"Why do you think you can count the same resistance twice? The current
only
goes through it once, regardless of whether it is being used as a motor
or a
generator."

He is not counting anything twice. Go back and read the post carefully.
You can rely on anything that Jim says as gospel.

Bob Swinney

In this case I'm not ready to grant Jim deity status. Perhaps you'd
better go back and read it twice, Bob. Here it is.

With an 80% efficient motor, 20% of the input power is lost in
armature resistance.

Now this is true.


Operated at the same speed as a generator, 80% of
the original motor input power is available as generated output.

This is at best misleading. There is no "original motor input power"
here. The motor isn't being used as a motor. It is being used as a
generator. There's *external* mechanical power driving the motor
shaft. This could be equal to 50% of the "original motor input power",
or it could be 80% of the "original motor input power", or it could be
100% of the "original motor input power", or it could be 200% of the
"original motor input power". In other words, it is *totally independent*
of whatever power the motor may have drawn when being used as
a motor.

However it now has to travel through the armature resistance before it
reaches the output terminals and this loses a further 20% of the
available power.

*Now* he can account for armature losses. If they are 20%, then
the generated output will be 20% less than the mechanical input
power applied to the generator.

Gary