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### Topic: JohnRob: AC Motor question (Read 359 times)previous topic - next topic

#### avr_fred

##### Oct 01, 2017, 06:22 pmLast Edit: Oct 01, 2017, 06:34 pm by avr_fred Reason: Removed unnecessary comment

Your original quote to which I took exception was "Keep in mind, motors work on current, not voltage.  The torque created is directly proportional to current though the windings."

The key phrase here is "directly proportional".

The definition of direct proportionality: "If one variable is always the product of the other and a constant, the two are said to be directly proportional. x and y are directly proportional if the ratio yx is constant."

Mathematically, y = kx and k = y/x with k being a constant.

Would you agree the above represents direct proportionality and the images below show this concept graphically?

#### LandonW

#1
##### Oct 01, 2017, 07:11 pm
an electric motor under load at slow speed requires more amps at less voltage and vise versa for when its at a higher speed.

#### JohnRob

#2
##### Oct 01, 2017, 08:38 pmLast Edit: Oct 01, 2017, 08:39 pm by JohnRob
@avr_fred,

Yes, I concur that the plots you attached meet the criteria for "directly proportional"

And yes, I concur that some of the input current goes to inducing current into the rotor.

But I still maintain that torque is directly proportional to current.   I though I would test our point of contention with real numbers so I went to myelectrical/tools/induction-motor-calculator.

Making the assumption that at constant excitation voltage    power = voltage * current.   I know the current has a real and imaginary part, however I going to assume that we are talking about the real part.

I ran the calculator through 4 different conditions (of power input) and recorded the resulting torque.

This is way I found.

#### MarkT

#3
##### Oct 01, 2017, 08:57 pmLast Edit: Oct 01, 2017, 09:02 pm by MarkT
an electric motor under load at slow speed requires more amps at less voltage and vise versa for when its at a higher speed.
For DC permanent magnet motors, torque is proportional to speed - also true for BLDC's using permanent
magnets on the rotor.  And this is gross torque, not the net torque seen after friction in the brushes or
bearings has been accounted for.

Other types of motor with a field winding, this rule doesn't apply.

Especially not with induction motors as the original thread was discussing - its much more complicated,
and torque depends on both the current in the windings, the rotor flux and the relative (geometrical) phase
difference between stator current and rotor flux phasors.

DC series wound motors are completely different to DC permanent magnet or DC shunt wound
motors too - they can accelerate to destruction if unloaded.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### JohnRob

#4
##### Oct 02, 2017, 12:10 am
Hi MarkT

The original discussion boils down to two view points:

1) Induction motor torque is directly proportional to current.  Was not stated but I was assuming gross torque.

2) Induction motor torque is not directly proportional to current due to rotor induction current.  ( I think I have that correct).

Neither point was directed to absolute torque and associated losses but the proportionality of torque to current.

The mention of rotor flux has come up in several posts.   I recall that rotor flux was the result of induced current in the rotor, and the induced current in the rotor is a function of stator flux.  Also in the mix is slip angle.

So my contention remains that torque in an induction motor is directly proportional to current.  Neither point suggested the absolute value of that proportionality.

As for DC motors,  I would agree that power output is proportional to speed (and torque), but I'm pretty sure torque alone is proportional to current.

Interesting discussion.

#### avr_fred

#5
##### Oct 02, 2017, 02:04 amLast Edit: Oct 02, 2017, 02:22 am by avr_fred Reason: typo
Quote
But I still maintain that torque is directly proportional to current.   I though I would test our point of contention with real numbers so I went to myelectrical/tools/induction-motor-calculator.

Making the assumption that at constant excitation voltage    power = voltage * current.   I know the current has a real and imaginary part, however I going to assume that we are talking about the real part.

I ran the calculator through 4 different conditions (of power input) and recorded the resulting torque.

This is way I found
What was the point of this exercise? You'll find the equation is:

T = P * 9549 / N

Where
T = Torque in Nm
P = power in kW
N = speed

Of course this satisfies your requirement, it has a constant in the equation. But, I don't see amperage mentioned anywhere.

Jumping ahead, you said in this thread: "I know the current has a real and imaginary part, however I going to assume that we are talking about the real part."

Yes, that was exactly what I was talking about. The real part as found in the real world. A clamp on ammeter works for me.

If you had a 1 kW (or 1hp, doesn't matter) motor at the ready, I'd suggest measuring the amperage with the motor unloaded. You'll find the current to be about 50% of the full load rating. Load the motor to its rated torque at rated speed and the current should match the nameplate full load rating. Plot those two points. This is what you'll see:

Torque is X, Current is Y. Now, satisfy the requirement of y = kx and k = y/x with k being a constant.

That's my very simplistic point. Overly broad statements of motor behavior can appear to be incorrect when compared to real world measurements.

I fully understand the reasons why the measurements appear as they do. I've been explaining it plant electricians for more years than I'd like to admit.

#### JohnRob

#6
##### Oct 02, 2017, 03:53 am
Hi avr_fred,

I think see your point,  unless I'm wrong you are saying the no load offset current causes the proportionality to not be "direct".   If you take that position mathematically you are correct.  However if your calculation treats that as an offset then adding additional current will directly increase torque.

Just a note:  If you put a clamp on ammeter on the motor power feed them measure the motor voltage and multiply them together you will not get watts, you will get VA.   To get watts you must use a watt meter (I have a Valhalla) or an oscilloscope and measure the phase angle and make the calculation of real watts.

I believe in the equation you cited "T = P * 9549 / N"  where P is in kW and I believe P is Volts * amps.  If volts are held constant i.e 120VAC then P would be proportional (here's that word again) to current.

So I think our disagreement is largely our view on whether or not proportional must be 0 based (i.e. start at 0,0) or not.

I do thank you for the discussion.  Its always enjoyable to have an gentlemanly discussion with an intelligent individual.

John

#### edgemoron

#7
##### Oct 02, 2017, 06:06 am
Are you forgetting AC power factor, which is low in an unloaded induction motor, that's why the current is high but real Watts are low.

#### MarkT

#8
##### Oct 02, 2017, 12:07 pm
So my contention remains that torque in an induction motor is directly proportional to current.  Neither point suggested the absolute value of that proportionality.
It may be your contention, but reality doesn't agree with you.  Within the low slip regime the torque
is proportional to slip, and the real part (resistive) of the current phasor also, but the imaginary (reactive)
current is not proportional to slip so that the rms current is not proportional to torque, but the power
flow is.

A lot of the graphs people show for induction motors show only the resistive part of the current phasor
against slip/speed, not the rms current (which does not go to zero at zero load, the motor looks like
a large inductor in fact).  The power factor against slip is the curve that shows this.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### JohnRob

#9
##### Oct 03, 2017, 02:38 am
Hi MarkT,

If I understand you correctly I should have said torque is directly proportional to the real component of  current.    Your probably correct.  However I was originally called out for my torque statement when trying to give the OP a sense that the ceiling motors he was asking about do not work in the same way he might be used to with "regular" AC motors.

John