Thanks !
In the spirit of not re-inventing the wheel
are you free to experiment with this thing? Simple stuff?
Why i ask is this
induction motor establishes a field in the rotor
which, upon removal of external power, dies out with the rotor's natural L/R time constant
on big motors that can be a goodly part of a second
and during that time that rotor field will induce voltage in the stator, which decreases in both magnitude and frequency as the field dies off and the rotor slows down.
The job of those huge capacitors is to extend the amount of time the rotor's field persists
which begs the question
by how much did they really need to extend rotor decay time? In other words, how much capacitance do they need?
We could embark on a research project, observe voltage decay on a motor and estimate rotor time constant that way
see
http://scholarworks.boisestate.edu/cgi/viewcontent.cgi?article=1050&context=electrical_facpubs&sei-redir=1&referer=http://www.bing.com/search?q=induction+motor+rotor+time+constant&qs=n&form=QBLH&pq=induction+motor+rotor+time+constant&sc=1-35&sp=-1&sk=&cvid=B2E246CDE9E84E089C19605491FC6377#search="induction motor rotor time constant"
but one experiment would be a lot easier.
At the instant brakes are applied
current will be :
(voltage made at motor terminals by the rotor field spinning inside the armature ) divided by (sum of impedances) ,
and sum of impedance is Zmotor + Z external
and our best measurement so far is your 29 amps though i suspect it's higher because of meter response time. We'll use it anyway.
If Vmotor is at that instant let's just guess 130 volts,
130/29 = 14.4 ohms is sum of motor and external impedance
We know the capacitors are only 1/2πfc = ##]frac{1}{377 X 1000uf}## = 2.65 ohms,
and motor winding resistance is typically only a couple ohms, throw in a little more for interconnecting wires
suggesting motor impedance dominates and it'll be inductive.
In fact since inductive and capacitive impedances subtract, motor Z might be around 16 ohms.
Do we reallly need all that capacitance to extend motor time constant? I don't know, the more capacitance you have the longer it can maintain terminal volts as it slows down and frequency drops off..
The calculations are complex
but
you have access to a test bed !
What If
we lifted this wire and observed whether it significantly affects coastdown ?
View attachment 103374
That leaves the capacitors in series,
if your 29 amps for ¼ second gets larger then we are moving closer to series resonance between motor and capacitors
if it gets smaller then motor impedance is way less than my ~14 to 16 ohms
but the real question is how does it affect coastdown ? Does the mechanism stop at the right place?
To try and explain (to myself as much as to you)
thinking in simplest terms
Given that the objective is to remove energy from the rotating parts so the mechanism doesn't overshoot its resting place,,, and
Capacitance does not remove energy, resistance does,,, and
The only resistance in the circuit is motor winding resistance (well plus Rs of the capacitors where we'd rather not deposit energy for it heats them)
maybe there's a different balance of capacitance and resistance we could use that avoids this high current.
Remember your power theorem, maximum transfer when Zload = conjugate of Zsource
if Zmotor = 2+j13 , maximum slowdown will be when Zload = 2-j13
and our load is now roughly 0-j2.65
lifting that lead leaves the caps in series making our load 0-j10.6 at 60 hz
so the current might get really big.
If it does, that's a clue
do you think we might extract energy faster with some resistance external to the motor?
Maybe a 300 watt incandescent lamp here ?
View attachment 103375
Trial and error, but based on fundamentals...
every piece of data improves our understanding of the fundamentals.
old jim