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Dynamic braking 3

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Sparweb

Aerospace
May 21, 2003
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CA
I'd like to check my understanding and calculations in the subject of electrically braking a 3-phase self-excited generator. The generator is driven by a wind turbine, and by shorting the leads, the increased current in the generator winding can bring the turbine blades to a stop. I have already put braking switches on some wind turbines, and they do come to an immediate stop - or an "immediate slow" for want of a better term, because the wind keeps them turning slowly when they have been shorted like this. If you prefer, this seems to be similar dynamically braking a motor, but rather than simply disconnecting the line power and inserting a resistor, in my case the lines are disconnected and then shorted together to bring it quickly to a stop.

Looking through some references such as the Baldor Cowern papers, I find some info that helps but nothing really addresses this use of a generator. I'm keenly aware that the current spikes as soon as you do this, but that's why I want to know how to model the situation. This is clearly hard on switch contacts, and I don't want the current spike going higher than the locked-rotor amp rating (in motor terms) in the windings, either. I don't have equipment that can measure the spike, but I think I can do the math on it.

My understanding is that at all times, there is always an EMF (electromotive force) proportional to the speed of the generator, so for simplicity I'll say that it's (240V/1800RPM)= 0.13 V/rpm. If this is somehow oversimplified I hope you can point me to a better way.

Because the leads are shorted, there's no voltage at the leads. As a result, there is no EI power. However, a current is still driven by the EMF, through the winding resistance alone, demanding an I^2R power. I believe there will be no reactive power because E is zero, and for what it's worth, power factor is 1.

If the above is true, then the shaft power & torque of the generator can be determined by I^2R power alone, and this is rather simple because the current is only EMF/R, where R is the line resistance through the motor winding. Putting this into practice:

EMF = (240V/1800RPM)= 0.13 V/rpm
R = 1.0 Ohm
I = EMF / R = 0.13 A/rpm

For 100 RPM:
I = 0.133 A/rpm * 100 rpm = 13.3 A
P = I^2 * R = (13.3 A)^2 * (1.0 ohm) = 177 Watt

For 1000 RPM:
I = 0.133 A/rpm * 1000 rpm = 133 A
P = I^2 * R = (133 A)^2 * (1.0 ohm) = 17.7 kW

This looks reasonable and it also seems to reflect what I've been doing. Throwing the shorting switch when a wind turbine is turning fast is a recipe for disaster, and the numbers bear it out. On the other hand, when the wind is slow, shorting the leads is a convenient way to stop it. I like to have the dynamics and gyroscopic behaviour under control before lowering the tower, for example.

Putting this concept to use, if it's accurate, then to answer my concern about excessive current in the winding, I should consider the maximum rating and use the locked-rotor current ramp rate to find the maximum speed where this generator can be shorted.

I_max = 60 Amps
I_max / (0.133 A/rpm) = 450 rpm

Again, this seems to make sense. I'd still like a second opinion, if anyone has been able to follow this.

If this really is "so far so good" then I can go to the next step of specifying resistors to moderate the initial current spike, before switching to the dead-short that holds it. With 2 stages I believe I will be able to stop the turbine from any speed.
 
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Have you considered adding high-power filament lamps as the dump? If it takes too much current it will increase resistance. When it cools it will be at a few ohms. A few thousand hours of lamp life should translate to a few decades of operation. If the filament fails it's just open circuit and you can see if it's working properly if the lamps are where they can be seen.

Alternatively - toaster heating elements. Tougher than lamps, though not as fun to watch.
 
Hi Sparweb

I always thought that generated emf’s in windings were dependant on the inductance of the winding which I believe is a property of the winding itself. See this link
I might well be wrong but I can’t see how you can attain the emf by dividing the voltage by the rpm, there are others here much better qualified than I to answer this so I will sit and wait and learn something no doubt👍😀

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
For a shorting device, look for tungsten rated switches. They are designed to withstand a surge of 10 x rated current.
EMF depends on magnetic field strength, number of turns, speed.
Don't confuse EMF with terminal voltage.
More later, heading out to a minor hockey game just now.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
You may consider a two stage system.
Step one: Insert a low value resistor to slow the generator.
Step two: Short out the resistor.
But, low external resistance encourages voltage collapse.
It may be well to skip the resistor stage.
Maximum current: EMF divided by circuit resistance.That is the internal resistance of the alternator plus any external resistance.
(The EMF will be approximately equal to terminal voltage in the free spinning alternator.)
Any inductive reactance will increase the circuit impedance and reduce the current.
You may measure the resistance with an Ohmmeter.
I suggest designing for peak Amps to get the least I[sup]2[/sup]T.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
If the above is true, then the shaft power & torque of the generator can be determined by I^2R power alone, and this is rather simple because the current is only EMF/R, where R is the line resistance through the motor winding. Putting this into practice:

EMF = (240V/1800RPM)= 0.13 V/rpm
R = 1.0 Ohm
I = EMF / R = 0.13 A/rpm
I'm not sure inductive effects are irrelevant. It has two effects which I will label below as first effect and second effect.

The stator has an ac component flowing in an inductive circuit. The sinusoidal portion of I could be lower than you calculated due to the inductive effects that you neglected. (Call this error in calculating ac current the "first effect"). Ignoring those inductive effects would be conservative (with respect to estimating the sinusoidal portion of the stator currents).

On the other hand neglecting inductance might be non-conservative from the standpoint that suddenly connecting an RL circuit to an ac source results in not only ac component but also a decaying DC component in the stator (call this decaying dc component the "second effect"). At least that's what I'm thinking based on analogy to suddenly connecting any R/L circuit to a power source.

To illustrate the second effect more, consider DOL start of induction motor:
Ignoring the decaying dc component you would estimate the starting current as
Istart_rms = Vrms / |Zlockedrotor|
Istart_pk = sqrt2*Vrms / Zlockedrotor​
But if you consider those decaying dc componets you can get up to a worst-case instantaneous current peak of:
ACTUAL Istart_pk = 2*sqrt2*Vrms / Zlockedrotor​

Ignoring this decaying dc component could result to up to a factor of 2 error in your peak current estimate and factor 4 error in peak power estimate (due to the second effect).

The peak may not be as relevant as the RMS over several cycles. If you make an ultra conservative assumption of no decay in the dc offset then we find an upper bound on the rms error of sqrt(3) error in rms current and factor of 3 error in associated power. The actual error in several-cycle rms from neglecting dc component would be somewhat lower than that because of course the dc does decay.

In the end you can neglect all the fancy calculations by just adding enough margin. I don't know what you will be using as a basis to determine your current limits and over what timeframe and how razor sharp the calculationi is to begin with. If you are looking at instantaneous peak currents, then a factor of 2 error in current estimate is conservative to account for second effect (dc offset). On the other hand if you are looking at several-cycle heat dissipation then sqrt3 margin into your current estimate would conservatively address that second effect of dc offset (with some extra margin from neglecting decay of dc offset). In both cases you have some additional (unquantified) margin due to that first effect (estimating current as I=V/R without considering series L).

Of course the total heat dissipated in circuit resistance (internal plus external resistance) over the entire duration of the braking would be equal to the initial kinetic energy of the rotating assembly IF there were no additional energy exchange between blades and air during braking (it'll probably higher due to that energy exchange). External resistance is the only way to move a portion of that energy generation outside of the winding. I'm pretty sure you know that piece since you mentioned external resistors.

I don't have any practical experience with dc braking (I'm sure Bill and others here do). I'm not sure I can add much more, but can you remind me if this is induction generator or sync generator, and single phase or 3 phase?


=====================================
(2B)+(2B)' ?
 
3DDave, you read my mind: I have some spools of nichrome, and this is exactly what I have it for.
waross, tungsten sounds like a specialty item, but I'll look it up nonetheless.

desertfox & waross, I admit I am displaying my ignorance by using the term "electromotive force" this way, but I know of no other term that describes the phenomenon that causes the potential to be generated, even if the terminal voltage is effectively zero because there is nearly no resistance between the two terminals. I could switch to say "open-circuit voltage" but that conceals the fact that it's linearly proportional to generator speed. You might still object to an expression of open-circuit voltage in units of Volt/rpm (or Volt/Hz if it matters). If it helps, you can ignore all references to variable speed of the generator, if you're more comfortable thinking about inductance as a constant at constant frequency. This machine is anything but constant frequency, but I'm used to that, so I'll work that out myself.

Electricpete, firstly, thank you for the in-depth analysis. Despite the uncertainties, you've identified more to consider than I could have, so I'll take some time to digest it. I may be able to gain some insight by looking at a model of the generator in normal operation, particularly if it narrows down the inductive component of the total impedance. Actually, now that you point it out, maybe I can work out "Z[sub]lockedrotor[/sub]" with the information I already have. Before installing this generator, I measured inductance across each phase with my multimeter and got about 40 milliHenrys. Can that value be used directly, or does it need some interpretation?

FYI, this is a synchronous 3-phase generator. The output is rectified to DC, but the dynamic braking switch is on the AC 3-phase leads. I expect the rectifier to have a negligible effect. Maybe it is also helpful to point out that the working speed range is 150 RPM to 500 RPM. This generator is speed-limited but not speed regulated.
 
electricpete said:
would be equal to the initial kinetic energy of the rotating assembly

Agree, the integral of the braking function over the time of braking is equal to the KE that is dissipated. Plus a bit because of the continued incoming wind power. The speed doesn't have to be reduced to zero for the integral to work. It is unavoidable that some of the energy is dissipated as heat in the stator. Improving my understanding of the braking energy may help me minimize that and maximize the amount in the resistors.
 
Not tungsten, "tungsten rated". The wall switches in your home are "tungsten rated".
EMF- The voltage generated, but divided between the load and the internal voltage drop.
With an efficient generator most of the EMF will be dropped across the load.
With shorted terminals, the EMF will be dropped across the generator winding.
With the generator stopped or creeping, the EMF will be very low.
Your interval of concern is the time to bring the rotor to a stopped condition.
This should be a very short time, possibly less than a second with a small generator with plastic blades, and the generator windings should be able to handle the heat of braking easily.
You will know best how quickly your particular generator may be stopped.
If you are able to arrange the shorting gear between the alternator and the diodes, you should be able to short the leads without shorting your battery.
Once you short the alternator output, the EMF will drop as the alternator slows, and as a result the current will drop also as the alternator slows.
You are familiar with the video of the powerful magnet dropped down a copper pipe.
You know how slowly it falls.
There are a lot of basic parallels between that affect and the stopping or very slow turning of a shorted, wind driven, permanent magnet alternator.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Spar Web said:
Improving my understanding of the braking energy may help me minimize that and maximize the amount in the resistors.
As I understand the driving forces on the propeller, there is not a lot of torque generated by the wind on a stopped blade.
Statically:
The less resistance in the circuit, the higher the current and the higher to braking force.
So we want as little resistance as possible in the circuit at rest.
Dynamically:
I am more concerned that it may be possible to damage the prop blades by stopping too abruptly than I am with electrical damage.
I may be in error.
E-Pete posted concerning possible DC offset when the alternator is shorted. I am just not sure if this will be a concern.
The DC offset will approach: RMS x (2 x √2) = 2.83
Use three times as an easy, conservative value.
So with the current calculated as RMS/R = I, three times that may be taken as a conservative approach value.
A relay or contactor intended for motor starting will safely withstand a starting surge of 6 times rated current.
Using a motor starting contactor that is rated for the normal full load current of the alternator as a shorting device should have a 200% safety margin.
That leaves the question of how rapidly you are able to stop your turbine without incurring mechanical damage.
I will stop when I am ahead and leave that question to you and others.
Was your carving machine useful in building the blades for your new wind turbine? (I hope so.)grin

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Oh, you mean a switch whose contact rating is suitable for lighting loads, including tungsten filaments. Gothca.
All the rest; we are on the same page, but I am still working on making the math better.

Just in case I didn't say so, I've been using a 30A 3-pole switch to short the AC leads of this wind turbine for 10 years, never had a problem. It does indeed take a second or two to brake from ~300RPM to about 20 when it's windy, ~150 to zero when the breeze is very light. My interest is to verify that I haven't just been "lucky" all this time. While replacing the switch, many questions came to mind, hence the investigation we have here. There is no reason to worry about torque on the blades I am using. They are overdesigned to a degree rarely seen on WT's.

By the way, yes the carving machine has made two matching blades, but not the ones being used here. The high price of lumber has suppressed my enthusiasm for wood carving, but only temporarily. That's why I haven't had much to show for that project, yet.
 
Hi electricpete,

Thanks again for your suggestions. I've tried them out and I have interesting results. I was tentative about including the reactive component but now that I've incorporated it this makes a lot more sense. Along the way I also found an interesting secondary effect but I may struggle to explain it clearly. A bit of important background is that this generator was built by taking apart a common 3-phase Baldor induction motor, removing the rotor's induction bars, and fitting a new rotor with permanent magnets. This conversion is what makes it a synchronous motor/generator, and allows it to self-excite at very low speeds.

Back to the math, I did have a measurement of the phase inductance (taken with a multimeter of unknown accuracy). Its vector sum with the resistance is the total impedance. Doing this, I was happy to see that the current levels off, which matches what I'd expect with a total circuit impedance that grows with increasing speed. Then I noticed that it actually levels off at over 60 Amps, which is the locked-rotor current spec of the original motor. Is that a coincidence or something of a foregone conclusion given that the generator in this project started life as a motor?

Anyway, I followed through on the rest of the analysis and found that the braking power increases rapidly at first, then becomes linear above about 300 RPM. This power curve is much steeper than the power curve acting as a generator, and also much higher than the incoming wind power at wind speeds corresponding to the rotor speed. All of this indicates that dynamic braking is effective, but not very effective. It's about 2X more torque. The current limiting property of the reactance prevents the heat from ramping up to infinity, and instead stays high but under some limit. I still haven't worked out if that self-limit is low enough to be safe, long-term, because the torque isn't high enough to really stop the turning.

If I extend the wind power curve out further, it ramps up with a cubic power, while the braking increases linearly, as I said. This means that the wind will eventually overcome the brake at about 500 RPM. This wind turbine has a separate speed-control mechanism that normally keeps RPM below 400 even in very strong winds. Coupled with the dynamic brake circuit with 2X more torque to overcome than normal, it may stay well below 300 RPM when dynamically braked. If I have the calculations right, then the current would go up to about 30 Amps in that condition. Maybe that's fine briefly, but too much for continuous operation. Probably worth noting that wind turbine generators receive a lot of "forced cooling" when they are working the hardest, by definition. But it may not be enough. It's also not a convincing case that the dynamic brake really prevents a runaway.
 
The locked rotor current may be six times the rated safe operating current.
At 400 RPM, the internal fan (If it is still in place) may not be moving much air through the end turns.
I firmly believe that the lowest resistance possible in the braking circuit will give the lowest braked speed.
This will result in the lowest wind power transmitted to the alternator and the lowest EMF and lowest current.
Apart from the difficulty installing the hardware, a shorting contactor adjacent to the alternator may be much more effective.
The length of the leads from the alternator to the controls may be limiting the braking force.

I recently worked on a Primus Air 30 turbine.
Link
This houses the controls and rectifiers in the alternator housing, and outputs DC on the slip rings.
Battery power is available to operate the electronics.
When the voltage rises above the set point (About 27 Volts with 24 Volt batteries) an internal circuit shorts the stator windings.
When the voltage reaches the set point, the turbine stops. Almost no drift.
We put a load on the batteries and the turbine started to support the load. Removed the load and the turbine stopped.



--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
To determine whether the amount of current is (long term) injurious, look at the time it takes to bring the winding to steady state temperature - which is different than the time to bring the whole mechanism to steady state. A reasonable rule of thumb is that the individual coils in the winding will reach steady state in roughly 300 seconds. Heating is going to be proportional to (current x current x time), where current is in per-unit and time is in seconds. Run a rolling average over the 300 second window: the sum of the (I x I x t) intervals should be under about 1.05 to be sure you're not damaging the winding. As an example: if you run 2 pu current for 60 seconds and 1 pu the rest of the time (within the 300 second window), the equivalent heat load in the coil is 1.6x rated ... which is far too high. A more reasonable heat load would be no more than 1.05x.

As you've discovered, dynamic braking is a high-energy event - hence having dynamic braking in industrial applications as a very short-duration occurrence.

Converting energy to motion for more than half a century
 
There are two applications of dynamic braking.
The first is retarding action, braking a locomotive descending a grade or holding a wind turbine at 400 RPM.
This is where Gr8blu's post is valuable. A similar issue is covered by Cowern under RMS Loading. (The Cowern Papers, on the Baldor Website)

When dynamic braking is used as a retarding action, runaway and over current are real concerns.
A second application of dynamic braking is braking to a (rapid) stop, and holding at a very low speed.
In this case, the final retarding force and the current will be much less and very much less than runaway values.
To gain a rough appreciation of the orders of magnitude, compare the power that the turbine is able to extract from the wind while turning at 400 RPM.
Compare that to the power that is able to be extracted at zero RPM or at the lowest speed that you are able to brake to.
While it may seem counter intuitive, As the combined resistance of the stator and the downleads to the shorting device is increased, the greater will be the fully braked speed and also the EMF and the current.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
Hi Gr8blu,

It took me a while to come back to this problem and see if I can apply what you suggested. It certainly makes sense, but also proves dynamic braking is trouble.

For any braked condition, the current will be high, but your criterion basically says that the average current must stay below the rated current. Accounting for the braking spike, the sustained current after that event has to be a little LOWER than the rated current.

In this turbine's generator, any rotation above 70 RPM will cause a short-circuit current of 8.9A which is higher than the motor rating, 8.8A. That only gets 100W of braking power, which is nowhere near enough to resist a strong wind. This puts numbers to my earlier hunch that this can't prevent runaway blades.

If the braking switch is closed at 400 RPM, as waross said, the current spikes to 40 Amps, but in that condition, the 50 kph wind driving the turbine blades would keep them turning at about 150 rpm, and the continuous 20+ Amps would eventually cause a breakdown.
 
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