Merlin3189 said:
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
Merlin3189 said:
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.
Merlin3189 said:
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.
lol
