Lenz’s Law Pictorial

Electrical & Electronic Engineering gen discussion Electric Motors & Generators
#1

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QUESTION

I need to roll out an electrical generation presentation to groups of Technicians to help underpin their understanding of generation principals. I was separately discussing with one of them about why Generators tend to slow down when load is applied, and how the Governor is increased to correct the frequency in Isoch mode etc etc. I was using the explanation of Lenz’s Law (along with Flemming’s Right Hand Rule and Maxwells Corkscrew Rule) for why this happens, but he was not understanding, and I ended up running out of ideas for ways of getting the phenomenon across to him, and he asked if I had any pictorials (which I didn’t).

I searched the internet for suitable pictorials (which speak a thousand words) but failed to find anything that clearly represented this, so my question is, does anyone have pictorials of this (not narratives – they are in abundance on the internet)?

I attempted to sketch it out myself, and found that it was more difficult that is first seemed, and actually led to some confusion on my part. I have attached my drawing (I’m sorry, yes, it’s pretty bad) and want to verify that what I am thinking is actually correct:

Lenz’s drawing:
https://www.sugarsync.com/pf/D7292559_79475805_6809935

It represents the Rotor outside the Stator so I could clearly show the Field Winding and Stator Winding directions, and I have shown just a single Stator Coil for clarity.

Using the sketch, this is my explanation:

  1. The direction of Field current around the salient pole heads means a North is on top and South is on the bottom, by way of Maxwell’s Corkscrew rule (MCR).
  2. This Field cuts the Stator coil and, using Flemming’s Right Hand Rule (FRHR), a Stator current is induced in the direction indicated with the arrow.
  3. Using MCR, this sets up a clockwise field around the Stator conductor (when viewed from the left).
  4. Changing to the small picture (left had side end on view), with the Rotor being driven clockwise by the Prime Mover, this clockwise Stator field interacts negatively with the rotating North Pole of the Rotor, and hence creates drag on it (tending to slow the Rotor down).

Could anyone please confirm or correct (probably the latter) what I am thinking, and also maybe provide some better pictorials that I could use.

REPLIES

electricpete

  1. This Field cuts the Stator coil and, using Flemming’s Right Hand Rule (FRHR), a Stator current is induced in the direction indicated with the arrow.

As small clarification, I’d prefer to break that into two steps:
2A – a voltage is induced in stator coil by Faraday’s law
2B – that voltage results in a current, depending on load connected. Assuming a resistive load, then the current is in phase with the voltage. And a little thought shows that the maximum current in the loop you showed occurs roughly at the time the rotor is in the position you show (when rate of change of flux through the loop is at its highest).

  1. Using MCR, this sets up a clockwise field around the Stator conductor (when viewed from the left).
  2. Changing to the small picture (left had side end on view), with the Rotor being driven clockwise by the Prime Mover, this clockwise Stator field interacts negatively with the rotating North Pole of the Rotor, and hence creates drag on it (tending to slow the Rotor down).

#4 leaves me flat. It is as you suggest interaction of rotor and stator fields that results in the force, but that can be tough to visualize. If you draw field lines it may give a hint (the field lines act like springs) but again not really intuitive to draw.

You could instead apply F = q V x B = i L x B to the stator conductors.
Look at the top stator conductor from the end view (lower right hand corner)
q V =i L is into the page
B is straight up
Resulting force is to the right. This falls out of the well known right-hand rule or cross product rule. Note that i must be defined as conventional current (if your audience thinks of electron current, it doesn’t work).

Similarly looking at bottom stator conductor from the end view (lower right hand corner)
q V =i L is out of the page
B is still straight up
Resulting force is to the Left.

The result of both of the above is a torque on the stator clockwise (in direction of rotation).

But assuming steady state condition (angular acceleration is zero), Newton tells us there must be an equal/opposite reaction on the rotor. The electromagnetic torque on the rotor is CCW (opposite direction of rotation). Mechanical power must be exerted to keep the rotor turning at constant speed against the electromagnetic torque. Perhaps we already knew that from conservation of energy: Neglecting losses in the machine, the amount of mechanical power in (w*T) input must equal the amount of electrical power out (I * V).

The idea of forces acting directly on conductor can be intuitive. But if conductors are embedded in slots, the torque producing force acts primarily on the iron core (not the copper conductors). I have some whitepapers that delve into that aspect. I’m sure it’s not an essential aspect for your audience, but some of the figures showing flux lines etc might help somewhere along the line.

http://electricpete1.tripod.com/torque_web/attach/LinkToTheLongVersion.htm

Here is a more direct link to the first one:
http://electricpete1.tripod.com/torque_web/attach/TheShortVersion091207.pdf

#2

I am just a lowly mechanical engineer. But, if I wanted to demonstrate Lenz’s law, I would drop a very powerful magnet down through a piece of copper pipe. I keep one in my desk for demonstration purposes.

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#3

My old powder scale had a copper plate working between two magnets for damping.

Dik

#4

Cool video but this is Eddy current loss not motor reluctance or load impedance regulation error.

#5

You are making it far too complicated. You can draw an analog with Ohm’s Law and Load regulation error spec on any power supply.

Load regulation error is defined as the drop in voltage from the rise in load current expressed as a % of the output voltage. Typically solid state regulators are 1 to 2%.

An audio amplifier uses the inverse of this ratio and calls it Damping Factor where 100 is a good value and very high end high power audio equipment may be up to 1000. That makes a an Audio Amp almost an ideal voltage source but not quite. Any real voltage source of either DC or AC has an output or source impedance as it is called, similar to an effective series resistance in a dielectric or battery which if you apply a step load current in either direction , results in a small step change in voltage from ESR then a ramped voltage due to capacitance. Little batteries like Li Ion have incredibly large capacitance and small ESR. The 10 watt-hour (Wh) 18650 Lithium Ion cell found in toys and e-bikes has more than 10 thousand Farads when new and fully charged (10kF), whereas a Maxwell double-electric layer Supercap of 10kF would cost more than $1kusd if you could find one. But then the Super Cap would have only 1 or 2 milliohms ESR and the 18650 cell might be 60 mΩ or less when fully charged and new and only last 250 to 1000 cycles depending how gentle they are treated.

This Damping Factor is not the same effect as a magnet inside a copper tube or over a block of aluminum. This viscous damping effect is drag effect of very high magnetic flux of rare earth magnets going thru a non-magnetic material that happens to have eddy current losses.

I digress

Getting back to Ohm’s Law and Generator loading.

A generator will have some MVA, KVA or VA rating and rated to support Vo / Io = Zo load impedance .
It will have some feedback loop to regulate the field current which acts as an electromagnetic AC field like a spinning clutch in an engine transmission except a clutch is designed not to slip under load. Whereas a cable, coil and magnetic mutual coupling field must have some losses (unless it was a cryogenic superconductor). Everything in this universe has some resistance or AC impedance.

The closed loop load regulation error based on an unregulated or open loop impedance of Z=ΔV/ΔI just as any cable has resistance. R=V/I but here we call it incremental resistance…

The grid is powered by generators and feeds Power Transformers which distributes the power to many stages of Distribution Transformers and ends up at a customer’s site in breaker panel. The grid in North American is designed with with a service goal of 10% line and load regulation tolerance for 120Vac 60 Hz where 5% is allowed for generation which is very dynamic with automatic tap changers from the source and then 5% for by fixed tap changers in the DT’s including few % for the cable from the last shared DT to service entry. But the generators can be controlled over the entire power range of power load at a steady RPM not only at same frequency but phase corrected to grid unless it is an island power source.

So my advice is define Lenz’s Law but don’t bother with the complicated picture. Just use an electromagnet to illustrate stiffness which means conductance or the inverse of Impedance. So by increasing the field current, the magnetic flux is stronger and the generator impedance is lower so it can supply more current while the RPM governor keeps it at a stable frequency with some rules and regulations defined by specs and regional power authority.

Give them a 1 minute history lesson on who contributed to magnetism theory ( mostly dozens of German and Swiss mathematicians) https://www.britannica.com/science/electromagnetic-radiation/Relation-between-electricity-and-magnetism#ref307404 before Tesla created AC generators to power the grid.

Sorry for the lack of editing and structure. I realize this is more like a conversation than a textbook answer of describing a car alternator on the grid and not a coal generator or nuclear power plant, but you get the idea, I am sure.

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#6

I can see why your student is confused.
The input power to the generator equals the output power plus the losses.
If you add load, you are asking for more power out.
But the power out is limited by the power in.
As a result the generator slows down.
The governor sees the slight drop in speed and increases the power in.
More water, more gasoline, more diesel fuel, more jet fuel, more steam, etc.
Cruise control in an automobile may be more applicable than Lenz’s law…
Look at control theory.
You only have to look as far as the proportional band of a PID controller to understand how the great majority of generators are controlled.
Even in utility generation and distributed generation all but the swing set may be in droop mode.
Droop is a type of proportional control. 3% droop equals 3% proportional band plus 3% offset.
For the very few governors in isochronous mode, include the Integral section of a PID controller.
Not much Lenz.

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#7

OK By definition, with 3% droop if the full-load speed is 100% and the no-load speed is 103% or the no-load speed was 100% then the full load speed is near 97%.

But why does it slow down with a load was the question I answered. That means if you were already applying full fuel energy, why does it slow down. Then the magnetic coupling impedance impedance matters as this relates Lenz’s Law to load regulation error.

#8

It’s 103% No Load and 100% full load.
Why do you have to push the accelerator further to maintain your speed when going up hill?
At no load and 103% speed, there is actually very little fuel being used.
As the load is applied the engine slows down, just as your car will slow down when climbing a hill if you don’t open the throttle.
As the engine slows down, the governor opens the throttle until at full load the engine is at full throttle and running at 100% speed.
Magnetically, with no load, there is theoretically no load on the engine. Actually there are parasitic loads, great and small.
At the low end is the excitation. Not much but it does demand some power to maintain.
Then there is windage and friction.
At the top of the losses by far is heat rejection.
Even at no load the governor has to supply enough fuel to support these losses.
Now we put a load on the alternator.
This causes a current in the stator windings.
This current causes a magnetic field which opposes the field that created it.
The more load, the stronger this field is.
The stronger the stator field is the harder it is to turn the rotor against this field.

As far as percentages. 3% proportional band would give 100% at no load and 97% at full load.
Add 3% offset to move those values up to 103% and 100%

#9

Ah, here’s the rub. Full speed at no load does not equate to full fuel energy.
As the set is loaded the fuel delivery is increased until you reach full load and full fuel delivery.

#10

That was already assumed. If you equate full speed at no load , there would be little fuel need to cover the no-load losses. Again that does nothing to explain the problem asked. All electric motors behave this way but do not have a 3% droop. It could be 50% droop but that is not acceptable since grid specs are 10% f.
Its a matter of source /load impedance which is related to torque,power, V/I and Lenz’s Law for magnetic impedance. The same is true for DC power supplies as source /load impedance ratio defines the drop in regulated voltage with load according to the loop gain on feedback error for %P droop then when the load exceeds supply limits, it saturates and falls out of linear regulation. In that DC case may not be a magnetic saturation problem, but could be. It’s usually a thermal current limit or a voltage drop limit.

#11

???
Well an standard squirrel cage induction motor will approach synchronous speed at no load, slowing approximately 3% for a 1740 RPM motor and approximately 2% for a 1760 RPM motor, at full load.