Spinning up a flywheel with a motor

[ag123]

the calcs are uploaded in the jupyter notebook
https://www.kaggle.com/ag1235/flywheel
shared publicly

 

[ag123]

the motor equation in this model is:

$$ i^2 R + K i^2 \omega = { I \omega ^ 2 \over 2 } + \text{heat from coil resistance} + \text{heat from mechanical friction} \ldots ( 1 )$$
$$ \text{where i is current, R is resistance, K is the inductance of the motor!} \\
\omega \text{ angular velocity, I is moment of inertia ... taken over 1 second. i.e. energy} $$

and

$$ K i^2 \omega = { I \omega ^ 2 \over 2 } \ldots ( 2 )$$
the LHS is energy stored in the inductor and RHS is actually the energy stored in the flywheel
https://en.wikipedia.org/wiki/Flywheel

heat from coil resistance + heat from mechanical friction is lumped into the i^2 . R resistance term in (1)
mostly only (2) is used as i^2.R is simply heat. due to w being present on both sides of the equation. the energy on the LHS is 1st calculated and used to calculate the flywheel energy on the RHS. i think the iterative sequence is literally a numerical integration of some sort. this analysis probably has applications beyond the simple flywheel

interestingly, if there is more heat generated, it would imply a lower w, and power = torque . w
which means torque would go up. But the catch with this model is w cannot be 0 or torque -> infinity.
i.e. this still can't model the stall torque. perhaps the stall torque is some psuedo w, i.e. w is some rpm when nothing is turning at all, real world is non-linear (plastic as magnetic flux is limited). so in the mean time, i'd just take a clip and hold the motor to measure the stall torque, no flywheel lol

 

[zoki85]

Total loses are difficult to calculate accurately for tiny motors. Only msms would give real picture.
Here is the graph max efficiency vs motor size for small "BLDC" and classical brushed DC permanent magnet motors:

Bellow m ≈ 1.3 g and output less than ≈1 W it is nearly impossible to attain 50% efficiency mark.
It should be emphasized that max efficiency does not occur at max power.

 

[Merlin3189]

the motor equation in this model is:

$$ i^2 R + K i^2 \omega = { I \omega ^ 2 \over 2 } + \text{heat from coil resistance} + \text{heat from mechanical friction} \ldots ( 1 )$$
$$ \text{where i is current, R is resistance, K is the inductance of the motor!} \\
\omega \text{ angular velocity, I is moment of inertia ... taken over 1 second. i.e. energy} $$

$$ K i^2 \omega = { I \omega ^ 2 \over 2 } \ldots ( 2 )$$
the LHS is energy stored in the inductor and RHS is actually the energy stored in the flywheel
https://en.wikipedia.org/wiki/Flywheel

heat from coil resistance + heat from mechanical friction is lumped into the i^2 . R resistance term in (1)
mostly only (2) is used as i^2.R is simply heat. due to w being present on both sides of the equation. the energy on the LHS is 1st calculated and used to calculate the flywheel energy on the RHS. ...

I'm puzzled by these equations. I presume all the terms are power (in Watts.)
I don't know what inductance we're talking about here, but I don't understand why the energy stored in it should be multipled by ω to get a power.
If the LHS is the electrical input power, I'd have thought the two terms would be the $i^2 R \ \text{ and } iV_{bemf} \ \ \$ being the power dissipated in winding resistance and the power used in the motor effect.
The $iV_{bemf} \ \ \text{ term would be equal to }\ \ i K ω \ \$ where K was the motor's back emf voltage constant in Volt per radian per second, rather than an inductance. (NB. you may come across various forms of voltage constant in common use, including an inverse , rpm per volt.)

interestingly, if there is more heat generated, it would imply a lower w, and power = torque . w
which means torque would go up. But the catch with this model is w cannot be 0 or torque -> infinity.

Certainly you are right about more energy going into $i^2R$ heating in the windings at low ω, because the current is inversely proportional to ω. Equally the torque will correlate with $i^2R$ heating, because torque is proportional to current.
But you seem to be going back to a constant power output.
Output Power = torque x ω is ok, but when ω goes to zero, so does power output.
Even if you had constant power input, at ω=0, all that power would simply go to the $i^2R$ term and more heat is generated.
Nowhere does torque or current go to infinity. Max current and max torque would ideally occur at ω=0, though heating of the windings raising their resistance would then reduce them a bit.

i.e. this still can't model the stall torque. perhaps the stall torque is some psuedo w, i.e. w is some rpm when nothing is turning at all, real world is non-linear (plastic as magnetic flux is limited). so in the mean time, i'd just take a clip and hold the motor to measure the stall torque, no flywheel lol

It models stall torque if you do it right.

If you measure stall torque, it doesn't matter whether there is a flywheel or not - it's not moving.

If you want to measure stall torque, you don't need to do it at operating voltage stall current, risking overheating the motor, causing damage and incorrect reading due to changed resistance. You can measure the stall torque at a lower current, by using a lower voltage supply (I'd say below half of stall current at working voltage.) You know torque is zero at zero current. Since torque is proportional to current, you can extrapolate to the torque at calculated stall current for the operating voltage.

 

[ag123]

I'm puzzled by these equations. I presume all the terms are power (in Watts.)
I don't know what inductance we're talking about here, but I don't understand why the energy stored in it should be multipled by ω to get a power.
If the LHS is the electrical input power, I'd have thought the two terms would be the $i^2 R \ \text{ and } iV_{bemf} \ \ \$ being the power dissipated in winding resistance and the power used in the motor effect.
The $iV_{bemf} \ \ \text{ term would be equal to }\ \ i K ω \ \$ where K was the motor's back emf voltage constant in Volt per radian per second, rather than an inductance. (NB. you may come across various forms of voltage constant in common use, including an inverse , rpm per volt.)

the equation actually equates energy

LHS is the electrical analogs of energy in each second, RHS is the physical mechanical (energy stored in spinning disc) and heat energy terms

the i^2 . R term is easy to understand it is simply heat generated by resistance. i place the same on the right as the heat generated from both heat generated from resistance + heat generated from mechanical friction

i initially had a hard time trying to figure out the analogs of the motor energy that translates to actual torque. it turns out that is the back emf, so the motor energy modeled as an inductor.
$$ \text{back emf} = L\,\frac{d i}{d t} = L i \omega\\
\text{power} = i v = L i^2 \omega $$ this models the energy stored in the motor 'inductor' for the particular second.

the physical mechanical analog on the RHS is the kinetic energy stored in the flywheel
$$ E = { I \omega ^2 \over 2 } $$ but the RHS is really the cumulative sum of the energy received from the motor each second and the disc speeds up.

i mean to say that the equation is really a recurrance expression, it is perhaps incorrectly expressed as I'm not sure how to express that well, perhaps i should say
$$ \sum _ { t = 0} ^ t { L {i _ t} ^2 \omega _ t } = { I { \omega _ t }^2 \over 2 } \dots (2) $$ initially i totally stumbled trying to figure it out as the current varies on the LHS and the back emf term depends on w (omega, angular velocity) and w changes every second as more energy is pumped into the flywheel. in the end i decided to use an iterative procedure in which i compute the back emf energy term in the first second add that to the total energy in the RHS and use the flywheel kinetic energy term to compute the new w (omega, angular velocity). next i take the new value of w and use that to compute the next iteration of the currents. since i hold the voltage constant (you can assume it is a buck converter holding the voltage supplied constant)

$$ Vin = i R + \text{back emf} = i _ t R + L i _ t \omega _ t $$ and it turns out it works ! The L term here is the K term in the original expression. that is in the jupyter notebook and accounts for these new curves.
https://www.kaggle.com/ag1235/flywheel

Certainly you are right about more energy going into $i^2R$ heating in the windings at low ω, because the current is inversely proportional to ω. Equally the torque will correlate with $i^2R$ heating, because torque is proportional to current.
But you seem to be going back to a constant power output.
Output Power = torque x ω is ok, but when ω goes to zero, so does power output.
Even if you had constant power input, at ω=0, all that power would simply go to the $i^2R$ term and more heat is generated.
Nowhere does torque or current go to infinity. Max current and max torque would ideally occur at ω=0, though heating of the windings raising their resistance would then reduce them a bit.

constant power output is not assumed here. rather the equivalence is between the back emf term of the motor and the kinetic energy gained in the flywheel. after i compute energy in the flywheel, i can get the angular velocity from the expression $$ E = { I \omega ^2 \over 2 } $$ then with the angular velocity i can compute torque from power = torque x angular velocity. the accidental low torque and constant power is a coincidence out of the calculations. the power turns out to be near constant i'd guess simply because the flywheel stores all that energy and speed up, this results in near constant or lower torque and power needed even though the flywheel gain speed.

It models stall torque if you do it right.

If you measure stall torque, it doesn't matter whether there is a flywheel or not - it's not moving.

If you want to measure stall torque, you don't need to do it at operating voltage stall current, risking overheating the motor, causing damage and incorrect reading due to changed resistance. You can measure the stall torque at a lower current, by using a lower voltage supply (I'd say below half of stall current at working voltage.) You know torque is zero at zero current. Since torque is proportional to current, you can extrapolate to the torque at calculated stall current for the operating voltage.

thanks i'd try that approach

another thought perhaps hopefully we can improve the equation / expression so that the stall torque can be modeled as well. it seem that it would be some kind of if - else expression as the stall torque occurs simply due to motor reaching its magnetic flux limits and couldn't produce more torque.

placing the angular velocity as zero in the expression gives absurd infinite torque due to the use of the expression power = torque x angular velocity. but literally in a sense the flywheel isn't gaining any energy as angular velocity = 0.

i'm also suspicious that there could be a friction term that increase with the angular velocity. the notion is that if the friction is independent of angular velocity and that the flywheel keep gaining kinetic energy, the rpm term will keep going up until the disc shatters. the term that currently offset it is in the back emf on the electrical analogue side, as the rpm increase back emf increase and current reduce and eventually taper out. but the curves gives much higher rpm than reality suggest and the rpm level off at a no load speed for small motor.
(edit: I'm suspecting for the motor itself, it may be due to the air friction drag within the motor itself
https://en.wikipedia.org/wiki/Skin_friction_drag
if this is true, a real disc may go on to spin up to rather high speeds, given in the open, there is much lower drag)

the expression seemed quite general but i think the other heat based terms are harder to express in other contexts than a simple flywheel. an example is that instead of a flywheel, the RHS is really a gear motor or rather a set of gears, those energy expressions may not be as simple to derive as a simple minded flywheel. and literally even the disc (flywheel) isn't needed, the motor rotor itself is the flywheel.

 

[sophiecentaur]

You have failed to understand post #2. DC motors are not 100% efficient. They do not go to infinite torque at zero RPM. They do not slip. They do not stall. They simply provide non-zero torque at zero RPM.

The commutator let's you down at low speeds but a brushless motor (mentioned only once above) has better low speed torque.

But the only way to have any real idea about what's going on is to measure V and I as they vary during the whole period of acceleration. This applies to motor car power systems too yet students are often given "SUVAT" style questions in the context of "A motor car accelerates from 10mph to ...". Very confusing.

Simulations such as the one suggested can be very misleading. It's always potentially GIGO.

 

[ag123]

omg, i found that this little virtual flywheel inertia dynamometer actually is able to predict some things.
the stall torque for my little motor is around 20 gf.cm !

how do i know? in my previous post the 2nd chart, there is a rather flatline portion on the torque v rpm graph reposted as follows:

that flatline portion is basically that from the berkeman's post

so by fitting a line and projecting it to 0 rpm, it gives a reading closer to 20 gf.cm
edit: nope incorrect conclusion, if i use a heavy flywheel, the chart changes.
what that means is that those flatline curves are one of the valid conditions
current model is not capable of predicting the stall torque, it needs to be measured

 

[Merlin3189]

placing the angular velocity as zero in the expression gives absurd infinite torque due to the use of the expression power = torque x angular velocity.

That's the point for me. If angular velocity is zero, then power is zero whatever the torque. So you can draw no conclusion about torque. In fact that's how stall torque is defined, the torque when angular velocity is zero.

Would you equally say that if the torque fell to zero, the angular velocity had to be infinite? Or again simply that it meant the power was now zero, whatever the speed. And the speed is now (nearly) the free running speed.

the equation actually equates energy

LHS is the electrical analogs of energy in each second, RHS is the physical mechanical (energy stored in spinning disc) and heat energy terms

So the equation actually equates energy in each second, otherwise known as power.
Which is also what $I^2R$ is

i initially had a hard time trying to figure out the analogs of the motor energy that translates to actual torque. it turns out that is the back emf, so the motor energy modeled as an inductor.
$$ \text{back emf} = L\,\frac{d i}{d t} = L i \omega\\ \text{power} = i v = L i^2 \omega $$
this models the energy stored in the motor 'inductor' for the particular second.

I would quibble with your definition of back emf here.
Back emf is not due (mainly) to inductance. The reason that back emf is proportional to speed, is that is caused by the dynamo effect of wires (the rotor) moving in a magnetic field. That gives you a relation
$v ∝ ω \ \text{ or }\ v=Kω \ \$ where K is the (operating voltage)/(ideal free-running speed).
You can work K out from windings and field flux, but you'd need to know details of motor innards to do that. Usually we (the users) either get it from data sheets or determine it by experiment.

There may be some inductance in the rotor windings and I guess that's one reason our simple model is a bit rough and ready. What it actually does, I think, is to slow the growth of current as the poles commutate. So at high speed the average current could be a bit below what is predicted by (V - Vbemf)/R
There could also be some more losses due to magnetising and un magnetising the rotor.
We accounted for some loss in non-ideal windings with the $i^2R$. We could add some for eddy currents and sparking, which I'd guess might go up with i and with ω. I can't remember seeing a treatment of this, though someone must have done it.
I assume that for simplicity we take it over most of the useful motor range, as a side effect subsumed in "frictional losses"
The useful power input is that represented by current x dynamo back emf.I have some sympathy with your difficulty in applying these relations. I've failed to come up with a function for ω(t) and have been hoping one of the mathologers here might mention it just to show us how clever they are.
I do have my own numerical simulations (very rough - in Excel) with a bit of inertia to give me an ω - t graph.
One thing it showed me immediately, something I should have realized at the start, is that the flywheel never reaches steady speed. It approaches it asymptotically. So you may have trouble timing it from 0 to steady state speed. You'd do better to pick a value around 60% and time it to that.

Now that I've satisfied myself that the ideas I've been pushing do indeed give me the textbook graphs, I think my next step is a few experiments with some nice little motors I have, to see if I can get the simulations to match up with reality.

 

[ag123]

I would quibble with your definition of back emf here.
Back emf is not due (mainly) to inductance. The reason that back emf is proportional to speed, is that is caused by the dynamo effect of wires (the rotor) moving in a magnetic field. That gives you a relation
$v ∝ ω \ \text{ or }\ v=Kω \ \$ where K is the (operating voltage)/(ideal free-running speed).
You can work K out from windings and field flux, but you'd need to know details of motor innards to do that. Usually we (the users) either get it from data sheets or determine it by experiment.

There may be some inductance in the rotor windings and I guess that's one reason our simple model is a bit rough and ready.
...
The useful power input is that represented by current x dynamo back emf.

I have some sympathy with your difficulty in applying these relations. I've failed to come up with a function for ω(t) and have been hoping one of the mathologers here might mention it just to show us how clever they are.
I do have my own numerical simulations (very rough - in Excel) with a bit of inertia to give me an ω - t graph.
One thing it showed me immediately, something I should have realized at the start, is that the flywheel never reaches steady speed. It approaches it asymptotically. So you may have trouble timing it from 0 to steady state speed. You'd do better to pick a value around 60% and time it to that.

Now that I've satisfied myself that the ideas I've been pushing do indeed give me the textbook graphs, I think my next step is a few experiments with some nice little motors I have, to see if I can get the simulations to match up with reality.

thanks for your response ! i had a hard time initially figuring out how to connect the electrical analogs on the electric side to the physical mechanical side. so in the initial attempt i only modeled just the flywheel. then as i researched further on it turns out one of the 'easier' way to represent a motor on the electric side is to represent it as a resistor and an inductor in series. the resistor part is trivial, the notion is it is simply heat and ignored. that 'inductor' is the interesting part accounts for the back emf and the torque generating components. I kind of postulated that as inductors store energy, this energy should be the energy that gets transferred to the flywheel and as this energy is not turned into heat, it has no where to go but simply spin up the disc (or even if there is no disc the motor rotor itself). i decided to use the full 'inductor' model to represent the energy on the electric side and make that equivalence to the physical mechanical flywheel side and it turns out it works !

then i noticed a problem nearer to the last few posts. my little motor has a no load rpm of 9000 rpm. i found it out by simply pasting a little flap of tape on the motor spindle and spin it up. then i simply recorded audio. I'm astonished that that is what it is 9000 rpm, it nearly hit the manufacturer's specs.
https://www.pololu.com/product/1117/faqs

but the catch is that if the motor 'inductor' keep pumping energy to the RHS - the flywheel. the speed of the flywheel will keep increasing. and in fact my graphs show that it did not taper out at the noload rpm of 9000.
so i figured that something is missing in the analysis. i think that missing thing is the skin friction drag
https://en.wikipedia.org/wiki/Skin_friction_drag
this i figured that there is skin friction drag in the motor as the motor spin up to a constant speed.
i actually improved my model (offline) by considering the energy transferred to be

energy from inductor back emf - skin friction drag = energy transferred to flywheel

and using some such analysis where skin friction drag (within the motor) is a function of w (angular velocity), I'm thus able to model the topping out, behavior at 9000 rpm as at that point, no further energy is added to the flywheel. the results considering skin friction drag looks as follows.
the rpm no longer keep rising with time but level off, it looks quite real.

so my model is currently still absent the stall torque.
but i think these models are after all rather accurate as they are derived from the basics, the resulting curves (e.g. torque-rpm) are anything but 'linear'. i'd guess overly complicated models makes it difficult to design things at operating speeds and torque. Hence the 'linear' models are commonly used. you can see in the torque rpm chart the right portion is rather 'flatline' this is quite a coincidence with those 'linear' model.
the stall torque portion can't be modeled in this process. i actually played with my model by using much larger flywheel and that 'flatline' portion no longer looked flatline and is somewhat a curve.