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Re: emf, potential, voltage



If I describe the force due to a wire L meters long carrying a
current i amps in a field of magnetic intensity B webers/m^2
as F = BiL newtons,
then its similarity to the voltage e opposing movement in the
direction of that force (often called 'back emf') is notable:
e = BvL volts
where B and L have the same meaning as before, and
v represents the conductor's speed across the flux in m/s

As usually expressed for a motor, torque is a radial distance
constant times force, and force is given by a term comprising
B and L called phi, the net flux per pole and armature
current i, thus
T = k phi i

Meanwhile, the back emf e is expressed as rotation speed n
times pole flux phi
e = k n phi

Comparing expressions for force F and torque T it is visible that
the scaling factor k represents (principally) the
effective armature conductor radius.

Comparing expressions for back emf e, the terms v velocity
and k n evidently represent k as (principally) the effective
armature conductor radius as well.

In Imperial or US customary units, the k scaling factors also
carry the burden of units conversion. I imagine that the
simplicity offered by SI is being referenced in the U Mich material.

I think of a force in newtons acting over a distance in meters
representing work in watts as a comparable illustration of the
SI's simplicity in these units.

Brian

At 21:28 3/16/01 -0500, Bob Sciamanda wrote:

Google turned up this URL
http://www.engin.umich.edu/group/ctm/examples/motor/motor.html

I'm just beginning to digest it, and right away I don't see the
justification for the last statement in this quote:

"The motor torque, T, is related to the armature current, i, by a constant
factor Kt. The back emf, e, is related to the rotational velocity by the
following equations:

Torque = Kt *Armature Current,
Back emf = Ke* Rotational Velocity

In SI units (which we will use), Kt (armature constant) is equal to Ke
(motor constant). "

Should this be obvious? Can someone explain?


Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
www.velocity.net/~trebor
----- Original Message -----
From: "John S. Denker" <jsd@MONMOUTH.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Friday, March 16, 2001 8:18 PM
Subject: Re: emf, potential, voltage


At 02:30 PM 3/16/01 -0500, Bob Sciamanda wrote:

Is the motor a linear device?

To first order, it's linear :-)

If I apply various voltages across the
motor and plot V vs I will I get a straight line?

There is an OTBE issue here (Other Things Being Equal).

--- If you hold the rotation rate constant, you should get a reasonably
decent approximation of a straight line. See below for more on this.

--- If you don't hold the rotation speed constant, you can get horribly
complicated results. For instance, if you have a stalled motor, applying
additional voltage may get it unstalled, resulting in *less* current.

For instance, does not
the motor type matter (eg series or shunt field winding)?

Why should it?

An unloaded motor is of little use.

Agreed.

Is not the mechanical load a
parameter which will surely affect the V/I curve?

Yes! That's exactly the point.

Given a point (V1, I1) on the I-V curve, it is *much* smarter to model it
as a back emf of V0 and an impedance of
Z := (V1-V0)/I1 (equation 1)
I1 = (V1-V0)/Z (equation 2)
rather than as no back emf and a larger impedance
V1/I1

In particular, changing the rotation speed will change the back emf in
equation 2, leaving Z unchanged to a good approximation.

Bottom line: equation 2, with a back emf that depends on rotation rate,
is
a very serviceable model for a motor.

-- Can anybody offer a simpler model that makes comparably good
predictions?
-- Can anybody offer a comparably simple model that makes better
predictions?


brian whatcott <inet@intellisys.net> Altus OK
Eureka!