DFIG Wind Turbines generators Star/Delta connection 2

[acerox]

Hi,

I have a question in regards DFIG (Double Fed Induction Generators) used in big wind turbines (2MW for example).

http://www.eng-tips.com/postedit.cfm?id=8035900&action=edit

Generally speaking those generators work slightly above the synchronous speed, then the converter allows a higher slip below or above that synchronous speed.
However they can also be in Delta or Star connection. Usually they star in Start connection and then switch to Delta. I have read that the reason is because Star connection allows lower cut-in rpm and Delta connection higher power (because the current going through the windings is not so high, 1.73 lower than in Star).

I thought that the cut-it rpm where define by the synchronous speed (plus converter), so frequency and number of poles, but not the connection, so I am missing something here…

Any help understanding it would be much appreciated.

Thanks.

 

[davidbeach]

If the field was being supplied with DC then what you're saying about synchronous speed would be true. But the field is being supplied with a AC voltage; synchronous speed can be anything the controller wants it to be.

 

[acerox]

Can you be more specific? If it is a 3 phases, 6 poles generator in a 60Hz network, the synchronous speed should is 60*60/3=1200 rpm. The generator will produce power at 1200-1220rpm (1-2% of slip). if the RPM are let´s say 200rpm the generator will not generate any power.

 

[davidbeach]

I don't know all the math, but you can apply the right frequency to the field so that synchronous speed can be below 200 rpm. There's also probably a gear box involved, but ignoring that, the whole point of a DFIG is to create a synchronous speed well below what it would be for an induction machine with same number of poles.

 

[acerox]

Yes, but you cannot change the slip so much, so for a traditional squirrel cage generator (with no converter) the slip is typically 1-2%, If you add a converter in the rotor side (and a DFIG), that slip is increase at around 30%, and if you have a full converter, as you said, then you can match whatever frequency needed.

So just focusing on the DFIG and a converter for the rotor, the slip can be up to 30%, so if the synchronous speed is 1800rpm, the connection could be done at 0.7*1800=1260rpm. What I don´t understand is the Star/Delta connection effect. I have seen turbines starting in Star connection (for the stator) and then switch to Delta, the reason seems to be because power can be produced at lower RPM with Star connection and I don´t get this.

 

[MatthewDB]

The voltage on the rotor increases with increasing slip. At zero speed, the DFIG is a transformer, with the rotor voltage equal to the line voltage times the rotor to stator winding ratio.

To save cost on the inverter, the inverter does not have the voltage rating sufficient to allow for delta connection at stall. By switching to wye, the inverter only needs to handle 57% of the full delta rotor voltage.

The extra loss of the higher winding current is more than offset by the reduced magnatizing loss. (Remember low speed means low power, the losses are dominated by iron loss)

 

[acerox]

So, what you are saying is that in star connection the voltage is lower (57%) and so the magnetization losses?

In some turbines the switching star/delta is not active, so they connect directly in delta, so I guess the inverter can handle that voltage.

Well then, what´s all about the synchronous speed? I see generators (DFIG) whose synchronous speed is at 1500rpm are connecting at 850rpm, which is 43% less, is the converter allowing such a big slip?

Sorry for keeping asking but my background is not electrical, so some of these concepts are not clear for me.

Thanks.

 

[MatthewDB]

You should think of a DFIG as a synchronous motor and not an induction generator. The difference between an DFIG and conventional synchronous generator is that unlike a conventional synchronous generator, the field of a DFIG has AC instead of DC exciting it. The name Doubly Fed Induction Generator is a bit misleading.

The machine itself is a wound rotor induction motor. In an induction motor (or generator), the rotor currents are induced by the stator. The rotor circuit is passive in the sense that it is only reacting to the stator flux and the slip. The rotor circuit is absorbing energy, either with a resistor or a energy recovery inverter. In an induction motor there is slip. It can only be a motor below synchronous or a generator above synchronous.

To make a DFIG out of a wound motor, the rotor is instead actively fed by a stiff voltage from the inverter. The flux produced by the applied rotor voltage summed with the rotor speed will always be at synchronous speed. So in your example of 850 RPM on a 1500 RPM DFIG, the inverter would produce 21 2/3 Hz so the 850 RPM rotor is summed with a 650 RPM rotor flux to get a air gap speed of 1500 RPM to match the stator flux. There is no "slip" because the to fluxes are locked into synchronism.

Whenever the speed is below synchronous the inverter is actually supplying energy to the rotor to sum with the speed of the rotor. That extra power required flows out the stator and is looped back through the inverter. At synchronous the inverter is only feeding DC, the energy is very low. Above synchronous the rotor is absorbing power and the inverter exports power to the grid.

As I alluded before, the inverter is very expensive so there is an attempt to make it as economical as possible. The way to do this is to minimize the current the inverter has to handle, and the way the current is minimized is to have as many turns on the rotor as the possible which means the same rotor flux can be produced by a lower inverter current.

I don't know what the voltages are on the machine of your example, but I'll throw out a hypothetical example. Let's say the stator voltage is 3.3kV, and the inverter runs at 690V on your 2MW machine. And further let's use the number of 1950 RPM as the maximum speed. At 1950 RPM, the machine exports 77% (1500/1950) of the power through the stator and 23% through the rotor and inverter. Now the lowest possible rotor and inverter current in that scenario is to have 690V at the rotor terminals at 1.3 times base speed. This is done by giving a rotor with 69.7% of the turns of the stator (690V/(1.3-1)=2400V; 2400V/3300V=0.697). Less turns on the rotor would mean more current and less than 690V at 1950 RPM. Above 1.3 times base speed the voltage would rise above 690V and the machine would have to shut down for over-voltage on the inverter.

Below the 1500 RPM base speed, the same limit would apply. Right at 1500 RPM, there is no (AC) rotor voltage (just the DC for the IR drop in the rotor). As the speed drops, the rotor voltage rises until at 0.7 of base speed or at 1050 RPM the inverter voltage is back at 690V again. Going any slower increases the voltage, until stall the full 2400V would be at the rotor. Since the inverter can only take 690V, operation below 1050 RPM isn't allowed. To prevent catastrophic failure of the inverter, line voltage must not be applied until the rotor is spinning at 1050 RPM.

Below the 1050 RPM point, the stator is disconnected from the line and shorted. In this configuration, the generator is now an induction machine, connected through the inverter to the line. Normally this is only used for startup to motor the turbine up to cut-in speed or to brake the turbine on shutdown.

The 'trick' of switching from delta to star extends the minimum speed lower. When the stator is in star connection, only 1/sqrt(3) or 57% of line voltage is applied to the each stator winding. The flux drops by 57%, and the matching rotor voltage drops by 57%. The rotor can slow down more until the inverter would hit rated voltage. In the hypothetical design, the stator would have 1905V per winding; at 1050 RPM the rotor voltage drops from 690V to 400V. The speed is no longer constrained at 1050 RPM for power operation. The speed can be further reduced to 0.4 of base speed or 606 RPM.

 

[itsmoked]

Wow. I have much more appreciation for wind driven DFIGs now. Thanks Matthew.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[acerox]

Thanks lot Matthew, great explanation. Let me put some numbers from my case to see if understood it.

2MW turbine
4 Poles generator in a 50Hz grid.
Stator Voltage 690V
Converter Voltage 480V
Rated gen rpm=1680rpm
Syncr speed =1500 rpm

So power through the converter is 1680/1500=10.71% (214kW).
The lowest possible rotor and inverter current in that scenario is to have 480V at the rotor terminals at 1.12 [1/(1-0.1071)] times base speed. Which done by giving the a rotor with 579.7% of the turns of the stator (480V/(1.12-1)=4000V, 4000V/690V=5.797)???

The minimum rpm would 1500*(100%-10.71%)=1339rpm, at that point the inverter would at 480V again.

By switching to Star connection, the voltage will drop 57.7%, so from 480V to 277V, and so the minimum speed from 1339rpm to 773rpm.

Would this be ok?

 

[MatthewDB]

Yes, you've got it. The only caveat would be that there needs to be some margin to handle a wind gust. When a gust hits, the pitch mechanism takes time to respond so the rotor will increase in speed. You wouldn't want to trip offline from a gust, so the maximum speed the rotor ratio is set for will be a bit higher. Figure something more like 1.25 - 1.3 times base speed.