In article , Gary Coffman says...

On Fri, 29 Aug 2003 00:51:51 +0100, wrote:
If we now run this machine as a generator AT THE SAME SPEED it
will generate the SAME 80v EMF. Note that this 80v is an open circuit
voltage which is the result of the rate at which armature conductors
cross the flux linkages, it is independent of the value of the
armature resistance. However it has to pass through the armature
resistance before reaching the output terminals At the maximum
permissible 1A load current, 20v will be lost leaving a net loaded
output voltage of only 60v = 60w i.e 60% of the rated input power of
the same machine used as a motor.

Right, but not 60% of its input power as a generator. Efficiency is still
80%, you've just arbitrarily limited the amount of input power you're
allowing to be applied to the generator input shaft. Efficiency is still
always the ratio of output power to input power, and now the input
power is at the shaft instead of at the winding terminals.

Yes and for his conditions the input shaft speed is indeed
specified as being the "SAME SPEED" as the motor speed on
the nameplate - he made that even clear enough that it sunk
in for me. So for his given boundary conditions the explaination
was quite enlightnening for me.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================