On Sat, 30 Aug 2003 04:30:57 +0100, wrote:

On Fri, 29 Aug 2003 22:49:30 GMT, (Gary Coffman)
wrote:

On 29 Aug 2003 10:30:24 -0700, jim rozen wrote:
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.

Yeah, I'd picked up on his same speed and same current (torque)
parameters. What I objected to was the statement that *efficiency*
as a generator was only 64%. That was using inputs and outputs
incorrectly, ie input as a motor and output as a generator. You can't
do that when calculating efficiency. It is always power out divided
by power in *at the same instant*. To use figures from two different
modes of operation doesn't give efficiency.

Note also that it isn't mandatory to follow his limits of same RPM
or same current either. Those were just his arbitrary choices.
Increasing either speed (voltage) or torque (current) will allow
greater input power, and greater output power, up to the thermal
limits of the motor/generator.

Gary


PLEASE read the original post carefully! The post did NOT state
that the efficiency as a generator was only 64%.

The postscript to the post commented that increased power could be
obtained by increased speed.

Increased current is not permissible because the load current is
chosent to be equal to the rated full load current when operating as a
motor. Currents higher than this will normally result in overheating.

Reading carefully is definitely helpful! Confusion arises from the
"same speed" assumption. I agree to the same-current constraint
but the speedlimit should be no-load speed which the motor presumably
can handle without damage. If the motor operated as a generator at
no-load speed then generated EMF is rated terminal voltage, terminal
voltage is that minus IR drop or 80%. We now have same current
hence same losses and same torque, same power in and power out
hence same efficiency, terminal voltage is 80% of nameplate rating
when run as a generator.

Both Gary and Jim allude to thermal limit hence max current. I
concur. They both also noted that spinning the motor faster as a
genny makes more volts (more power in, more power out) while loss is
proportional only to current. In this case it would actually be
*more* efficient, the situation being analogous to operating a motor
at 20% above rated voltage with same torque and current. In Jim's
example, if it were spun 20% faster than no-load speed then it would
produce rated terminal voltage at rated current. I doubt if 20%
overspeed would present a problem other than reducing brush and
bearing life a bit.