I have been trying to work this out from basic principles but the answers
are at best foggy:

Consider two electric motors (commutator) with identical inputs, say 250W,
and identical RPM, say 15,000. What are the factors that determine the motor
efficiency and thus the output power?

If I understand the physics, the output power is directly related to torque
if rpm are held constant. What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one of
the factors?

Thanks,

--
Michael Koblic,
Campbell River, BC

"Michael Koblic" wrote in message
...
I have been trying to work this out from basic principles but the answers
are at best foggy:

Consider two electric motors (commutator) with identical inputs, say 250W,
and identical RPM, say 15,000. What are the factors that determine the
motor efficiency and thus the output power?

If I understand the physics, the output power is directly related to
torque if rpm are held constant. What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one
of the factors?

Thanks,

--
Michael Koblic,
Campbell River, BC

Yes, if you think of the rotor as a type of lever being moved
by a megnetic field.

Best Regards
Tom.

"Michael Koblic" wrote in message
...
I have been trying to work this out from basic principles but the answers
are at best foggy:

Consider two electric motors (commutator) with identical inputs, say 250W,
and identical RPM, say 15,000. What are the factors that determine the
motor efficiency and thus the output power?

If I understand the physics, the output power is directly related to
torque if rpm are held constant. What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one
of the factors?

Thanks,

short answer "yes"

longer answer - you are looking for factors that affect losses, for your
example, they include resistive losses in the windings, magnetic losses due
to the air gap, bearing losses, air friction of the rotating parts,
hysteresis loss in the core, and so on. Armature diameter will affect these
factors in some way, so yes, it is a factor. It's just not a direct
relationship
--
Michael Koblic,
Campbell River, BC

Your question is vague. Output power = force times distance divided by time. Motor characteristics are a product of the type and
there are many. Generally the efficiency of a motor can be determined by the temperature rise in use. The hotter the motor gets,
the less efficient it is. However when using AC motors, the power factor should be considered as well, as it will identify
electric energy drawn, but not used. I think the motor you are identifying is a DC motor. Some DC motors will run on DC or AC.
They make their own AC with the commutator and brushes, like electric drill and vacuum cleaner motors. I think you should ask the
root question you have and give us specifics.
Steve

"Michael Koblic" wrote in message ...
I have been trying to work this out from basic principles but the answers are at best foggy:

Consider two electric motors (commutator) with identical inputs, say 250W, and identical RPM, say 15,000. What are the factors
that determine the motor efficiency and thus the output power?

If I understand the physics, the output power is directly related to torque if rpm are held constant. What then, in the nature
of the motor construction, will increase its torque? Is the diameter of the rotor one of the factors?

Thanks,

--
Michael Koblic,
Campbell River, BC

On Thu, 2 Sep 2010 21:23:45 -0700, "Michael Koblic"
wrote:

I have been trying to work this out from basic principles but the answers
are at best foggy:

Consider two electric motors (commutator) with identical inputs, say 250W,
and identical RPM, say 15,000. What are the factors that determine the motor
efficiency and thus the output power?

For given input power, efficiency is determined by losses: friction,
windage, current*resistance, hysteresis and eddy current.

If I understand the physics, the output power is directly related to torque
if rpm are held constant. What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one of
the factors?

Torque is the integral of vector products of magnetic field density,
armature current, # of armature conductors carrying that current, and
armature conductor radius.

On 09/02/2010 09:23 PM, Michael Koblic wrote:
I have been trying to work this out from basic principles but the
answers are at best foggy:

Consider two electric motors (commutator) with identical inputs, say
250W, and identical RPM, say 15,000. What are the factors that determine
the motor efficiency and thus the output power?

If I understand the physics, the output power is directly related to
torque if rpm are held constant. What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one
of the factors?

From the outside, the motor efficiency is (useful output power) /
(input power). The rest is lost, usually as heat (hopefully not as
noise, but you never know).

Your question about increasing the torque is misleading, though: you
don't want to know "what increases the torque" so much as "what
increases the _available_ torque at my desired output power level".
Diameter of the rotor is, indeed, one of the factors.

As an example of this, in the model airplane world there are two basic
choices in brushless motors: "regular" inrunner motors, that put the
field magnet on the motor shaft inside of the windings, and "outrunner"
motors that put the field magnets on the motor housing, which spins with
the motor shaft. Outrunner motors generate more torque at lower speeds
-- and lower efficiencies -- than inrunner motors. Why use them?
Because they don't require gearboxes, and as a system they're lighter
and more efficient.

So the question you wanted to ask was "what are the loss mechanisms?".
The answers that I know a

* Windage. The motor stirs the air, and maybe even has a built-in fan.
That takes power away from the output shaft.

* Friction. Bearings. 'nuff said.

* Hysteresis losses. Those coils are wound on iron, and it takes a
certain amount of energy to reverse the field in a piece of iron. That
energy heats up the laminations, but does you no good. Good magnet
steel has lower hysteresis, and good motor design puts enough of it in
that the hysteresis loss is lower. Poor motor design...

* Eddy currents. The iron those coils are wound on are conductive, and
the coils induce currents in the iron. The current takes energy to
establish, and swirls around and generates heat. Thinner laminations in
the armature reduces the effect, but thinner laminations increase cost
and decrease the amount of iron you can pack into a given space -- which
increases the hysteresis loss.

* Resistive losses. The armature is wound with wire (probably copper).
The motor current has to travel through this wire, which is resistive
and has losses. More wire takes more space, and requires more iron.
Better wire costs more.

Anything that you can do to reduce one of the above effects either costs
money, takes time to design, makes the motor bigger and/or heavier
and/or less robust, or makes some other effect worse.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

On 09/02/2010 11:23 PM, Michael Koblic wrote:
I have been trying to work this out from basic principles but the
answers are at best foggy:

Consider two electric motors (commutator) with identical inputs, say
250W, and identical RPM, say 15,000. What are the factors that determine
the motor efficiency and thus the output power?

If I understand the physics, the output power is directly related to
torque if rpm are held constant. What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one
of the factors?
Iron losses in the rotor are proportional to speed and field flux.
There is going to be an optimum for any particular output rating where
iron losses hit a minimum, and then gradually increase above that size.
Copper losses require a specific cross sectional area for minimal
resistance, but use too much copper and you increase eddy current
losses. These are probably the main factors. You need enough field
amp-turns to resist the armature field and maintain the field flux.
But, you want to keep the field flux low enough to keep the iron losses
down. So, there are a bunch of minima that are loosely related. You
want to juggle all these so that overall minimum hits near the bottom
of all the individual minima.

Jon

On Sep 2, 9:23*pm, "Michael Koblic" wrote:
I have been trying to work this out from basic principles but the answers
are at best foggy:

Consider two electric motors (commutator) with identical inputs, say 250W,
and identical RPM, say 15,000. What are the factors that determine the motor
efficiency and thus the output power?

Output power == torque x RPM
... within a constant, because RPM (revolutions per minute)
is the traditional measure, but you really want radians per second

If I understand the physics, the output power is directly related to torque
if rpm are held constant.

Proportional, not just related, by definition.

What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one of
the factors?

That's a completely unrelated issue. The torque results from magnetic
attraction, and your magnetized materials (the steel) have limits to
their magnetizability. You will need more steel, or steel capable
of higher saturation magnetization, to raise this torque limit, and
some very efficient flux coupling (tight fit of rotor/stator ) to
get the full benefit. There's also remagnetization of the materials
occurring, so that the motor doesn't just lock up but actually
continues to move... so the duty cycle makes a part of your steel
ineffective because its magnetization is swinging through zero.

Yes, also rotor diameter changes the torque (the force
times the radius is the torque, increase the radius and keep the
force constant... increase the torque).

The motor is likely to burn up if the saturation magnetization limits
are reached. It can also burn up because too much current
(in the resistive wiring) is required to reach the target
magnetization.
It can also burn up if the magnetization is changed too often (there's
heat generated due to remagnetization).

On Fri, 03 Sep 2010 14:05:38 -0500, Jon Elson
wrote:

On 09/02/2010 11:23 PM, Michael Koblic wrote:
I have been trying to work this out from basic principles but the
answers are at best foggy:

Consider two electric motors (commutator) with identical inputs, say
250W, and identical RPM, say 15,000. What are the factors that determine
the motor efficiency and thus the output power?

If I understand the physics, the output power is directly related to
torque if rpm are held constant. What then, in the nature of the motor
construction, will increase its torque? Is the diameter of the rotor one
of the factors?
Iron losses in the rotor are proportional to speed and field flux.
There is going to be an optimum for any particular output rating where
iron losses hit a minimum, and then gradually increase above that size.
Copper losses require a specific cross sectional area for minimal
resistance, but use too much copper and you increase eddy current
losses. These are probably the main factors. You need enough field
amp-turns to resist the armature field and maintain the field flux.
But, you want to keep the field flux low enough to keep the iron losses
down. So, there are a bunch of minima that are loosely related. You
want to juggle all these so that overall minimum hits near the bottom
of all the individual minima.

Jon
To answer the question, yes- rotor diameter has a direct effect on
torque. As does feild strength.

"Michael Koblic" wrote in message
...
I have been trying to work this out from basic principles but the answers
are at best foggy:

[...]

Thank you all. I knew it was not going to be simple.
Mo' better motor = mo' money. Got it!

--
Michael Koblic,
Campbell River, BC

Don Foreman wrote:


For given input power, efficiency is determined by losses: friction,
windage, current*resistance, hysteresis and eddy current.


So if someone needed a very efficent motor, what would they get using silver for the
windings? A couple percent?

Wes

On Fri, 03 Sep 2010 22:06:50 -0400, Wes
wrote:

Don Foreman wrote:


For given input power, efficiency is determined by losses: friction,
windage, current*resistance, hysteresis and eddy current.


So if someone needed a very efficent motor, what would they get using silver for the
windings? A couple percent?

Wes

About that. Resistivity of copper is 1.67, silver is 1.59.

Increasing magnetic flux density increases efficiency. This can be
done with better magnetic materials, tighter tolerances resulting in
shorter air gaps, rare earth magnets in PM motors, etc.

The usual tradeoff is efficiency vs cost. Size, weight and cost can be
traded for efficiency.

Tim Wescott writes:


Anything that you can do to reduce one of the above effects either costs
money, takes time to design, makes the motor bigger and/or heavier
and/or less robust, or makes some other effect worse.


Tim nailed it. It's TANSTAAFL.

You need to balance all those factors, as varied by your hours/year of use,
% of full load, $/KWH, PF charges, etc.

I recall the consulting engineer talking about spec'ing some new
1600-2000 HP pipeline motors [4160V].

He worked it out 5 ways, and then said
'....and here on, it is $$ for efficiency
...you can buy a better motor or more KWH...'

If it's a main drive motor on a carrier, the cost is ignored, but
efficiency, serviceability, and service factor [can it take 125% rated load
for 8 hours, if needed?] are king.

But at Spacely Sprockets, other issues are...

--
A host is a host from coast to
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

On Sep 16, 1:06*am, David Lesher wrote:
...
If it's a main drive motor on a carrier, the cost is ignored, but
efficiency, serviceability, and service factor [can it take 125% rated load
for 8 hours, if needed?] are king.

But at Spacely Sprockets, other issues are...


Swiss aircraft carriers must have the world's best motors.
Quality motors for model carriers:
http://www.maxonmotorusa.com/

jsw